GIFT   OF 
MICHAEL  REESE 


Cec^nical 


ELEMENTS 


OF 


MECHANICAL    DRAWING. 

USE    OF    INSTRUMENTS, 
GEOMETRICAL    PROBLEMS,    AND    PROJECTION. 


BY 

GARDNER    C.   ANTHONY,   A.M. 

» • 

PROFESSOR    OF    DRAWING    IX    TUFTS    COLLEGE;     DEAN    OF    THE    BROMFIELD- PEARSON    SCHOOL; 
MEMBER     OF     AMERICAN     SOCIETY     OF     MECHANICAL     KNt.IM  U.-. 


UNIVERSITY 


BOSTON,   U.S.A.: 
D.   C.    HEATH    AND    CO.,    PUBLISHERS. 

1895. 


Copyright,  1889,  189$, 
BY   GARDNER  C.  ANTHONY. 


PREFACE. 


IN   the  preparation  of  this  book,  and  others  of  the  TECHNICAL  DRAWING  SERIES,  it  is  the  aim  of 
the  author  to  provide  text-books  rather  than  copy-books ;  treatises  in  which  principles  should 
be  established,  and  methods  suggested,  but  freedom  permitted  in  their  application.     While  the 
treatment  of  much  of  the  subject  is  new,  it  has  long  since  passed  the  experimental  stage,  and 
nothing  is  here  published  that  practical  experience  in  the  class-room  has  not  justified. 

The  THIRD  ANGLE  OP  PROJECTION  is  used  exclusively,  as  is  required  in  all  good  practice,  and 
the  terms  TOP  and  FRONT  VIEW  have  been  substituted  for  PLAN  and  ELEVATION  by  reason  of  the 
confusion  arising  from  the  use  of  the  latter.  The  methods  employed  for  the  representation  of 
objects  oblique  to  the  planes  of  projection  have  been  found  to  give  a  clear  and  comprehensive 
understanding  of  a  subject  which  is  usually  much  encumbered  with  rules,  soon  to  be  forgotten. 
The  introduction  of  a  graphic  statement  of  problems  relieves  the  instructor  of  the  mechanical, 
and  frequently  laborious,  task  of  devising  suitable  problems,  with  proper  dimensions,  and  enables 
the  student  to  begin  the  drawing  without  delay.  The  treatise  on  Screw-Threads  and  Bolts  is  the 
embodiment  of  instruction  and  problems  given  to  classes  during  the  past  six  years,  and  is  an 
introduction  to  the  study  of  machine  drawing,  as  treated  in  the  second  book  of  the  series. 


4  .  PREFACE. 

It  is  intended  that  the  student  should  first  thoroughly  master  the  principles,  and  then,  un- 
aided, apply  them  to  the  solution  of  the  problems,  receiving  such  instruction  as  his  special  case 
may  demand.  By  this  means,  individual  instruction  may  be  given  to  large  classes,  and  the  energy 
and  talent  of  the  teacher  directed  to  the  giving  of  instruction,  instead  of  performing  the  petty 
details  of  devising  problems,  etc.  It  also  enables  the  more  competent  student  to  advance,  inde- 
pendent of  the  progress  of  those  who  may  require  more  time. 

This  system  has  been  successfully  applied  by  the  author  and  others  to  the  teaching  of  evening 
drawing-schools,  high  and  manual  training  schools,  and  college  classes. 

GAKDNEK   C.  ANTHONY. 
TUFTS  COLLEGE,  MASS., 

Sept.  15,  1894. 


CONTENTS. 


- 


i. 

THE  OUTFIT 9 

THE  USE  OF  INSTRUMENTS    .     .          10 

Preparation  of  the  Paper,  10 ;  The  Pencil,  10 ;  The  Scale,  10 ;  A  Study  of  Straight  Lines,  11 ; 
Practice  in  the  Use  of  Triangles,  12;  Use  of  Dividers,  13;  Inking,  14;  Lettering,  16;  The 
Drawing  of  Circular  Arcs,  17  ;  The  Compasses,  17  ;  Inking  of  Circular  Arcs,  18. 

II. 

GEOMETRICAL  DEFINITIONS  AND  USEFUL  PROPOSITIONS 20 

Points,  Lines,  and  Angles,  20  ;  Triangles,  21 ;  Quadrilaterals,  22  ;  Polygons,  23 ;  Circles,  23 ; 
Solids,  24;  Pyramids,  25;  Prisms,  25;  Cylinders,  26;  Cones,  26;  Sphere,  27. 

GEOMETRICAL  PROBLEMS 27 

General  Instruction,  27 ;  Perpendiculars,  28 ;  Angles,  29 ;  Triangles,  31 ;  Tangents,  32 ;  In- 
scribed and  Circumscribed  Polygons,  34 ;  Construction  of  the  Hexagon,  35 ;  Inscribed  and 
Circumscribed  Circles,  37. 

EXAMPLES 39 

Perpendiculars,  39 ;  Angles,  39 ;  Triangles,  40  ;  Tangents,  40 ;  Inscribed  and  Circumscribed 
Polygons,  41 ;  Hexagons,  41 ;  Inscribed  and  Circumscribed  Circles,  42. 

III. 
CONIC  SECTIONS , 43 

The  Ellipse,  44 ;  The  Inking  of  Curves,  47 ;  The  Parabola,  48 ;  The  Hyperbola,  49. 
EXAMPLES   .  50 


0  CONTENTS. 

IV. 
PROJECTION .51 

General  Instruction  for  Drawing,  55;    Problems  1  to  12,  56;    Objects  Oblique  to  Planes,  58; 
Problems  13  to  17,  59;  Special  Methods  for  the  Revolution  of  Lines,  Surfaces,  and  Solids,  61. 
SPECIAL  PROBLEMS  IN  PROJECTION „ .65 

V. 

THE  DEVELOPMENT  OF  SURFACES 68 

Problems  18  to  25,  70;  Development  of  Surfaces  of  Revolution,  72;  The  Cylinder,  72;  The 
Cone,  73;  Problems  26  to  28,  75. 

VI. 

THE  INTERSECTION  OF  SURFACES 76 

Problems  29  to  31,  77;   Use  of  Auxiliary  Planes,  78;  Problems  32  to  36,  79. 

VII. 
SCREW-THREADS  AND  BOLT-HEADS 81 

Spirals,  81;  Involutes,  82;  The  Helix,  82;  Problems  37  and  38,  83;  V  and  Squnre  Screw- 
Threads,  .84;  Problem '39,  86  ;  Conventional  Threads,  86;  The  Double  Thread,  86;  U.  S. 
Standard  V  Thread,  87  ;  Problem  40,  88. 

BOLTS 88 

Sphere  and  Cutting  Planes,  88  ;  Hexagonal  Heads  and  Nuts,  89  ;  Problem  41,  91. 

VIII. 
ISOMETRIC  AND  OBLIQUE  PROJECTION 92 


MECHANICAL    DRAWING. 


THE    OUTFIT. 

following  list  comprises  the  smallest  equipment  of  tools  consistent  with  good  work. 
J-     Their  selection  should  be  intrusted  to  one  experienced  in  their  use. 

6  IN.  COMPASSES,  joint  in  both  legs,  needle  points,  pencil  and  pen  attachment. 

5  IN.  DIVIDERS.     3  IN.  Bow  PEN.    3  IN.  Bow  PENCIL.     3  IN.  Bow  DIVIDERS. 

5  IN.  DRAWING  PEN,  provided  with  ebony  handle,  and  no  joint  to  blades. 

12  IN.  BOXWOOD  SCALE,  divided  into  16th  and  32nd  as  shown  on  page  10. 

21  IN.  "I"  SQUARE.  7  IN.  45°  TRIANGLE.  10  IN.  60°  TRIANGLE.  The  T  Square  may 
be  of  pearwood  and  with  a  fixed  head,  but  celluloid  or  hard  rubber  are  more  suit- 
able for  the  Triangles. 

HHHHHH  SIBERIAN  LEADS  for  use  in  compasses  and  bow  pencil. 

HHHH  SIBERIAN  LEAD  PENCIL  and  MEDIUM  HARD  PENCIL. 

HARD  RUBBER  for  ink  erasing.     SOFT  RUBBER  for  pencil  erasing. 

PEN  HOLDER.     FINE  PEN.     PEN  WIPER. 

BED,  BLUE,  and  BLACK  DRAWING  INK. 

ONE-OUNCE  TACKS,  either  iron  or  copper. 

SCROLLS,  three  or  four  varieties  having  curves  of  long  radii. 

PENCIL  SHARPENER.     This  is  readily  made  by  gluing  No.  0  sandpaper  on  one  side  of 
a  thin  strip  of  wood  6  in.  long  and  1  in.  wide. 

1C  x  21  IN.  DRAWING  BOARD. 


10  USE    OF    INSTKUMENTS. 

THE    USE    OF    INSTRUMENTS. 

PREPARATION  OP  THE  PAPER.  —  Place  the  drawing  board  with  long  edge  next  the  body.  Fasten 
the  paper  within  about  3  in.  of  lower  and  left  hand  edges,  observing  that  the  lower  edge  of  the 
paper  should  be  square  with  the  board.  Use  four  tacks,  one  at  each  corner.  The  tack  heads 
must  be  forced  flush  with  surface  of  paper  so  as  not  to  interfere  with  freedom  in  the  use  of 
the  T  square.  See  that  the  pencil  is  properly  sharpened,  as  much  of  the  accuracy  of  tlie  work 
will  be  dependent  on  the  care  used  in  keeping  it  always  in  order. 

To  sharpen  a  pencil,  remove  the  wood  from  both  ends  by  means  of  a  knife,  exposing  at  least 
§  in.  of  lead.  One  end  should  then  be  filed  to  a  round  point,  and  the  other  flat  or  to  a  chisel  edge, 
about  3*2  in.  wide,  —  a  pencil  sharpener  or  file  being  used  for  removing  the  lead.  See  Fig.  2, 
PI.  1.  The  size  of  paper  best  adapted  for  the  problems  of  this  book  is  11  x  15  in.  It  may  be 
obtained  from  an  imperial  sheet,  22  x  30,  by  cutting  the  latter  into  four  equal  parts.  The  problems 
being  designed  to  occupy  a  space  not  greater  than  10  x  14  in.,  a  margin  of  \  in.  will  remain. 

Find  the  centre  of  the  sheet  by  placing  edge  of  T  square  to  coincide  with  the  opposite 
diagonal  corners  of  the  paper,  and  draw  short  fine  lines  which  will  intersect  at  the  centre  C, 

Fig,  1,  PL  1.  To  the  right  and  left  of  this  point  lay  off  Tin., 
and  above  and  below  it  lay  off  5  in.  In  laying  off  dimensions 
from  the  scale,  do  it  by  means  of  the  round  point  of  the  pencil, 
and  make  sure  that  the  point  be  exactly  opposite  the  required 


lilt.  '///////>/  division  on  the  scale.     The  best  type  of  scale  for  general  draw- 

ing is  that  shown  by  the  accompanying  figure.     It  should  be 

made  of  boxwood,  with  a  white  edge,  and  being  graduated  as  in  the  illustration  may  be  conveniently 
used  for  scales  of  Full,  Half,  Quarter,  and  Eighth  size.     The  scale  should  never  be  used  as  a  ruler. 


USE    OF    INSTRUMENTS.  11 

Next  place  the  head  of  the  T  square  against  the  left-hand  edge  of  the  board,  which  will  be 
known  as  the  working  edge,  and  under  no  consideration  should  any  other  be  employed.  Only 
the  upper  edge  of  the  blade  should  be  used.  This  is  designed  for  the  drawing  of  all  horizontal 
lines,  and  by  sliding  the  T  square  head  upon  the  working  edge  of  the  board,  a  series  of  parallel 
horizontal  lines  may  be  drawn.  The  T  square  should  be  moved  by  the  head  only. 

Through  the  points  previously  laid  off  above  and  below  the  centre,  draw  horizontal  lines, 
using  the  pencil  against  edge  of  T  square  as  follows  :  — 

Hold  the  pencil  as  nearly  vertical  as  possible,  with  the  flat  side  of  the  chisel  point  pressed 
lightly  against  the  T  square,  and  draw  lines  from  left  to  right. 

Vertical  lines  are  to  be  drawn  by  means  of  the  triangle  used  in  connection  with  the  T  square. 
Fig.  1,  PI.  1. 

With  60°  triangle  in  the  position  shown  in  Fig.  1,  draw  vertical  lines  through  the  first-named 
points,  using  the  same  care  in  holding  the  pencil,  and  drawing  the  lines  from  bottom  to  top.  If 
the  triangle  be  too  short  to  draw  the  entire  line  at  one  stroke,  move  the  T  square  and  triangle 
until  the  desired  length  is  obtained.  In  general,  the  pencil  should  be  moved  from  the  body. 

A    STUDY    OF    STRAIGHT    LINES. 

PLATE  2. 

It  should  be  observed  that  the  only  benefit  to  be  derived  from  the  making  of  this  and  the 
following  drawings  will  be  in  gaining  a  knowledge  of  the  use  of  instruments.  If,  however,  the 
directions  given  for  performing  this  work  are  not  carefully  observed,  and  the  prescribed  methods 
closely  followed,  much  of  the  value  of  this  study  may  be  lost. 

Having  prepared  the  sheet  as  directed,  divide  it  according  to  PI.  2,  making  3J  in.  squares,  and 
the  vertical  distance  between  squares,  1£  in. 


12 


USE    OF    INSTRUMENTS. 


Fig.  9.  Divide  A  B  and  B  D  into  quarter-inches  by  means  of  the  scale,  and  through  the 
divisions  on  A  B  draw  horizontal  lines  by  means  of  the  T  square.  Through  points  on  B  D  draw 
vertical  lines  by  means  of  triangle  on  T  square,  as  in  Fig.  1,  PI.  1,  observing  carefully  the  in- 
struction for  use  of  pencil.  Afterwards,  measure  the  distance  between  the  lines  at  their  inter- 
section with  A  C  and  C  D.  They  should  be  exactly  \  in.  apart.  Great  care  should  be  used  in 
the  pencilling  of  all  drawings,  since  it  is  an  almost  universal  experience  that  the  inking  of  the 
drawing  will  be  little  or  no  better  than  the  pencilling. 

Fig.  10.  PRACTICE  IN  THE  USE  OF  TRIANGLES.  —  Divide  D  F  into  quarter-inches,  and  from 
these  points  draw  vertical  lines  as  follows :  Place  the  (10°  triangle  parallel  to,  and  nearly  coin- 
ciding with,  D  F,  as  in  Fig.  3,  PL  1.  Now 
place  the  45°  triangle  against  short  edge  of 
60°  triangle,  and  on  this  edge  slide  the  60° 
triangle  parallel  to  D  F,  until  it  is  about  |  in. 
distant  from  it ;  then  holding  the  60°  triangle 
firmly,  place  the  45°  triangle  in  position  shown 
by  the  dotted  lines  in  Fig.  3,  and  also  by  the 
accompanying  figure.  Use  the  60°  triangle  as 
a  base  on  which  to  slide  the  45°  triangle,  and 
draw  the  required  vertical  lines.  Draw  lines 
parallel  to  D  E  with  45°  triangle.  Lines  parallel  to  C  F  will  be  drawn  by  45°  triangle  used  on 
T  square,  as  in  Fig.  1.  Draw  all  lines  from  left  to  right,  when  possible.  Figs.  10,  12,  14, 
and  18  may  to  advantage  be  first  made  on  a  practice-sheet  which  may  serve  later  for  a  trial-sheet 
in  inking. 

Fig.  11.     Divide  lines  F  E  and  F  H  into  half-inches,  and  by  means  of  T  square  and  triangle 


USE    OF    INSTRUMENTS.  ±O 

draw  the  small  squares.  Subdivide  the  divisions  on  E  F  and  F  H  into  eighth-inches  and  draw 
the  intermediate  lines.  See  that  the  lines  start  and  stop  exactly  at  the  required  points. 

Fig.  12.  Use  the  divisions  already  obtained  on  G  H,  and  divide  H  L  into  half-inches :  draw 
lines  converging  to  a  point  as  shown. 

Fig.  13.  Draw  horizontal  lines  one-quarter  inch  apart,  using  care  to  hold  the  pencil  vertically. 
This  figure  is  designed  chiefly  for  an  inking  practice. 

Figs.  14  and  15.  These  squares  are  first  to  be  subdivided  into  other  squares  as  indicated  by 
the  dotted  lines.  This  will  facilitate  the  drawing  of  the  figures  by  making  the  use  of  a  scale  un- 
necessary, for  it  will  be  observed  that  the  outlines  of  the  figures  are  drawn  to  the  corners  of  the 
squares  save  in  a  few  cases  where  the  middle  point  of  the  line  is  used.  The  divisions  of  the 
squares  may  be  made  by  the  scale,  and  the  lines  drawn  full  instead  of  dotted.  In  pencilling, 
always  use  one  width  of  line  and  let  it  be  fine,  since  it'  is  only  in  the  inking  of  drawings  that  the 
distinction  between  light  and  shade  lines  is  made.  Do  not  pencil  the  lines  drawn  across  the 
figures  ;  these  are  known  as  section  lines,  and  are  to  be  drawn  in  ink  only. 

Figs.  16  and  17.  Divide  these  figures  into  25  squares  each.  As  the  divisions  of  the  lines 
cannot  readily  be  obtained  by  means  of  a  scale,  used  in  the  ordinary  manner,  observe  the 
following  instructions  : 

USE  OP  DIVIDERS.  —  To  divide  a  line  into  any  number  of  equal  parts,  when  the  number  of 
divisions  required  cannot  be  directly  obtained  from  the  scale,  proceed  as  follows :  Suppose  it  is 
desired  to  divide  the  line  A  B,  Fig.  4,  PI.  1,  into  5  equal  parts.  Open  the  dividers  to  a  distance 
equal  to  A  1,  about  one-fifth  of  the  required  space,  and  holding  them  at  the  joint  by  the  thumb, 
first,  and  second  fingers,  place  one  point  at  extremity  A,  of  line  to  be  divided,  the  other  point 
being  at  1.  By  rotating  the  instrument  alternately  in  opposite  directions,  as  if  describing  a 
series  of  semicircles  (the  path  of  the  point  being  shown  by  the  dotted  lincs^ii*-  Fig.  d^PI.  1),  lay 

E       J-'" 

OF  THE 


14  USE    OF    INSTRUMENTS. 

off  divisions  A-l.  1-2,  2-3,  3-4,  4-5.  The  point  5  being  beyond  the  extremity  of  the  line,  the 
divisions  are  too  great,  and  must  be  diminished  by  an  amount  equal  to  one-fifth  of  B  5.  Make  a 
second  or  third  trial  if  necessary,  so  that  the  last  division  shall  fall  on  point  B.  The  following 
method  may  also  be  used.  Suppose  line  A  B,  Fig.  4,  to  be  2|  in.  long,  and  it  is  required  to  be 
divided  into  5  equal  parts.  Place  zero  of  the  scale  to  coincide  with  point  A,  and  lay  off  by  scale, 
five  equal  divisions,  as  i,  1,  1|,  2,  2|.  From  the  last  point  C,  by  means  of  triangles,  draw  line 
B  C,  and  through  the  remaining  points,  draw  lines  parallel  to  B  C,  intersecting  A  B  ;  these  will 
divide  the  line  A  B  into  the  required  number  of  equal  parts.  The  dividers  may  also  be  used  in 
place  of  scale  to  set  off  any  equal  divisions  on  A  C.  In  using  dividers,  take  care  not  to  puncture 
the  paper  more  than  is  necessary  to  make  the  point  visible. 

Fig.  18.  Divide  the  left-hand  vertical  line  into  quarter-inches,  and  draw  horizontal  lines  by 
means  of  T  square.  These  are  to  be  used  as  a  special  inking  practice  and  may  be  pencilled  by 
full  lines. 

INKING.  —  Before  beginning  to  ink,  see  that  the  pen  is  clean  and  properly  sharpened.1  Having 
slightly  opened  the  nibs,  fill  the  pen  by  means  of  a  quill,  or  writing  pen,  that  may  be  inserted 
between  the  blades  ;  and  although  this  should  not  necessitate  the  wiping  of  the  outside,  care 
should  be  used  to  see  that  it  is  perfectly  clean.  Do  not  overload  the  pen  with  ink.  Having 
filled  the  pen,  nearly  close  the  nibs  and  try  the  width  of  the  line  on  a  piece  of  paper,  opening  or 
closing  the  nibs  by  means  of  the  screw  to  vary  the  width  of  the  line.  The  pen  should  be  held 
vertically  by  the  thumb,  first  and  second  fingers,  as  shown  by  the  figure  on  page  12,  the  thumb- 
screw being  held  from  the  body  so  as  to  be  .readily  adjusted  by  means  of  thumb  and  second 
finger.  Like  the  pencil,  the  pen  should  always  be  moved  from  left  to  right,  and  from  bottom  to 
top  of  board. 

1  Instruction  concerning  this  should  he  given  by  the  teacher 


USE    OF    INSTRUMENTS  15 

When  inking,  both  nibs  must  rest  evenly  on  the  paper,  and  the  pen  should  be  pressed  lightly 
against  the  T  square,  so  as  not  to  close  the  nibs,  and  thus  vary  the  width  of  the  line.  Never  ink 
backwards  on  a  line.  In  case  the  ink  should  not  flow  freely,  touch  the  pen  to  the  finger,  and  if  it 
fails  to  flow,  clean  thoroughly  and  refill.  Never  refill  the  pen  or  lay  it  aside  without  cleaning. 

Ink  Figs.  9, 10, 11,  and  12,  using  the  same  method  as  was  employed  in  pencilling  the  lines.  In 
Fig.  12,  allow  each  line  to  dry  before  inking  the  following. 

Corrections  in  inking  should  be  made  by  means  of  a  hard  rubber,  but  never  by  scratching  with 
a  knife.  In  using  an  eraser  of  any  kind,  do  not  exert  much  pressure,  for  it  does  the  work  no 
more  rapidly,  and  is  liable  to  roughen  the  paper. 

Never  ink  any  portion  of  a  drawing  until  the  pencilling  is  completed. 

In  inking  the  long  fine  lines  shown  in  Fig.  13,  go  over  each  line  twice  in  succession,  without 
moving  the  T  square,  endeavoring  not  to  widen  the  line  at  the  second  inking.  See  that  the  pen 
always  contains  ink  enough  to  finish  a  line,  as  it  is  difficult  to  continue  with  the  same  width  of 
line  after  refilling. 

The  width  of  lines  on  different  drawings  may  vary  slightly,  according  to  the  character  of  the 
work,  but  guard  against  too  fine  a  line,  which  is  an  error  common  to  most  students  and  many 
draughtsmen.  Remember  that  a  drawing  is  made  to  be  read.  The  skill  in  inking  does  not 
depend  on  the  fineness  of  the  line,  but  on  its  clearness. 

Two  widths  of  lines  are  used  in  Figs.  14, 15, 16,  and  17,  the  wider  being  known  as  a  shade  line 
and  its  use  fully  explained  on  page  55.  First  ink  all  fine  lines  and  afterwards  the  shade  lines, 
the  latter  being  as  wide  as  the  widest  in  Fig.  18.  The  section  lines  are  to  be  drawn  last  and 
always  without  pencilling,  the  space  between  the  lines  being  dependent  on  the  size  of  the  figure. 
In  this  case  they  may  be  made  about  equal  to  the  space  between  the  vertical  lines  of  Fig.  8,  PJ.  1. 
Do  not  ink  the  divisions  of  the  square  shown  by  the  dotted  lines. 


16  USE    OF    INSTRUMENTS. 

The  lines  in  Fig.  18  should  be  inked  in  the  manner  shown.  The  first  two  being  clotted  lines, 
the  third  and  fourth,  broken  lines,  and  the  last,  a  very  heavy  line.  Always  use  the  greatest 
possible  care  in  inking  the  dotted  lines.  Never  make  dots  any  longer  than  shown  in  Fig.  18,  and 
see  that  they  are  equally  spaced. 

Lettering.  —  The  subject  of  lettering  is  of  such  importance  to  the  mechanical  draughtsman, 
that  he  should  early  adopt  some  clear  type  and  acquire  proficiency  in  the  freehand  rendering  of 
it.  While  it  may  be  necessary  at  times  to  make  use  of  instruments  and  mechanical  aids  for  the 
construction  of  letters  and  figures,  they  must  usually  be  written  freehand  and  with  fluency. 
Many  good  drawings  have  been  disfigured  by  the  lettering  or  figuring  when  a  small  amount  of 
practice  would  have  enabled  the  draughtsman  to  execute  the  same  neatly  and  legibly.  Plate  10 
illustrates  the  two  types  of  Alphabet  recommended  to  students.  These  are  such  as  will  always 
be  acceptable  in  the  regular  practice  of  draughting,  and  their  study  will  afford  not  only  an 
excellent  freehand  exercise,  but  skill  in  neatly  figuring  and  lettering  drawings.  Both  types  sbould 
be  written  without  the  aid  of  instruments.  The  first,  or  Gothic,  is  the  more  simple,  and  should 
be  used  by  most  students.  It  can  be  quite  easily  written  by  means  of  a  hard-wood  stick,  sharp- 
ened to  a  point  like  a  pencil,  the  size  of  the  point  being  varied  according  to  the  desired  widtli  of 
the  line.  In  using  this,  care  should  be  taken  that  the  ink  be  black  and  slightly  thick.  The  small 
letters  should  be  about  two-thirds  of  the  height  of  the  initial  letters,  capitals  being  used  for  both. 
The  second  type  requires  considerable  practice,  but  being  very  clear  and  beautiful,  should  be 
mastered  by  every  practising  draughtsman.  The  student  is  recommended  to  practice  the  latter 
without  the  use  of  shade  lines. 


USE    OF    INSTRUMENTS.  17 

THE    DRAWING   OP   CIRCLES   AND    CIRCULAR   ARCS. 

PLATE  3. 

To  prepare  the  compasses  for  use,  remove  the  movable  leg  and  place  in  it  a  HHHHHH  lead, 
which  must  be  sharpened  to  .a  chisel-point  according  to  directions  given  for  the  sharpening  of 
lead  pencils.  Next  fit  the  leg  into  the  instrument,  making  sure  to  firmly  clamp  the  same,  then 
observe  first,  that  the  needle  point  and  lead  be  of  the  same  length  ;  if  not,  adjust  lead  to  needle 
point ;  second,  that  the  point  of  the  lead  stands  directly  across  the  instrument,  so  that  a  line  no 
wider  than  the  point  may  be  drawn.  The  lower  part  of  the  needle  point  leg  should  be  kept 
nearly  vertical,  so  that  the  point  will  not  make  a  hole  in  the  paper  while  revolving.  It  is  not 
essential  that  the  pencil  point  be  kept  vertical. 

The  compasses  are  to  be  held  at  the  joint  by  means  of  the  thumb,  first,  and  second  fingers, 
and  the  instrument  always  rotated  from  left  to  right,  clockwise.  In  placing  the  needle  point  on 
a  particular  centre,  the  compasses  may  be  steadied  by  lightly  touching  the  needle  point  by  a  finger 
of  the  left  hand.  The  line  should  be  stopped  as  soon  as  the  circumference  is  complete,  otherwise 
it  is  liable  to  be  widened.  This  is  very  important  to  observe  when  inking. 

Fig.  19.  Draw  the  fine  lines  passing  through  the  centre  of  the  figure,  using  full  lines  for  this 
purpose.  Their  position  is  indicated  by  the  dimension  lines,  which  latter  are  not  to  be  drawn. 
The  number  of  lines  required  by  the  problems  differ  slightly  from  that  shown  in  the  figures.  The 
diameter  of  each  figure  will  be  4  in.  Divide  the  diameter  into  quarter-inches,  and  draw  a  series 
of  concentric  circles,  using  care  that  they  pass  exactly  through  the  quarter-inch  divisions. 

Fig.  20.  A  practice-sheet  should-first  be  made  of  Figs.  20,  21,  and  24.  Divide  the  horizontal 
diameter  into  half  inches,  and  draw  a  series  of  circles  through  these  points,  tangent  to  the  large 
circle  at  A.  Use  care  to  have  each  circle  pass  through  A  and  the  required  points. 

2 


18  USE    OF    INSTRUMENTS. 

Fig.  21.  Divide  the  horizontal  diameter  as  before,  and  draw  semi-circles  in  the  following 
order,  —A  B,  H  K,  A  C,  G  K,  A  D,  F  K,  A  E,  E  K,  A  F,  D  K,  etc.  See  that  the  circular  arcs 
make  continuous  lines  and  pass  through  points  A  and  K.  Do  not  mind  the  position  of  the 
centre  so  long  as  it  be  upon  the  line  A  K,  and  enables  you  to  draw  a  semi-circle  through  the 
required  points. 

Fig.  22.  On  the  diameter  E  F  lay  off  quarter-inches,  and  construct  squares  as  follows  :  With 
45°  triangle,  draw  lines  A  C  and  B  D,  through  the  centre,  and  through  points  E,  0,  H,  F,  etc., 
draw  perpendiculars.  From  their  intersection  with  the  diagonals,  draw  horizontals  A  B,  K  L. 
etc.,  making  no  square  less  than  1J  in. 

Next,  set  compasses  to  draw  a  circle  of  f  in.  radius.  Do  this  by  laving  off  the  amount  on  the 
paper,  and  set  the  compasses  to  it ;  but  under  no  consideration  take  this  directly  from  the  scale 
by  means  of  the  instrument.  In  the  corner  of  each  square,  draw  a  circular  arc,  just  touching" 
but  not  intersecting  the  sides  of  the  square.  To  clearly  indicate  the  position  of  a  centre  which 
is  to  be  used  a  second  time,  lightly  pencil  a  circle  about  it  as  at  E,  Fig.  24,  but  never  put  the  point 
of  a  pencil  in  the  centre  to  enlarge  or  blacken  it. 

Fig.  23.  Divide  the  diameter  into  half-inches,  and  describe  circles  as  in  Fig.  19.  Do  not 
shade  these  lines  in  pencilling. 

Fig.  24.     Divide  the  diameter  into  half-inches  and  construct  similar  to  Fig.  22. 

INKING  CIRCLES  AND  CIRCULAR  ARCS. — In  inking  with  compasses,  use  care  to  have  the  pen, 
as  well  as  needle  point,  always  vertical.  The  directions  for  cleaning  and  filling  compass  pen  are 
the  same  as  for  ruling  pen. 

Ink  figures  19,  20,  and  21,  but  in  the  last  two  use  care  not  to  draw  a  second  line  until  the 
first  be  dry,  as  a  large  number  of  lines  pass  through  the  point  A,  and  would  cause  a  blot.  Alter- 
nate circles  in  Fig.  19  should  be  dotted. 


USE    OF    INSTRUMENTS.  19 

Always  ink  small  circles  first.  Circular  arcs  should  always  be  inked  before  straight  lines. 
Therefore  in  inking  Figs.  22  and  24,  put  in  circular  arcs  first,  using  care  to  draw  no  more  than 
a  quarter  of  a  circle,  and  afterwards,  ink  straight  lines  by  means  of  a  ruling  pen,  joining  the  ends 
of  the  circular  arcs. 

In  Figs.  23  and  24,  the  method  of  shading  circles  is  shown.  A  and  C  in  each  figure  represent 
parts  of  a  solid  ;  B  and  D  the  space.  Do  not  ink  the  shaded  circles  without  the  following 
preliminary  practice. 

Having  drawn  a  circle,  immediately  remove  the  point  of  the  compasses  from  the  centre  to  a 
position  slightly  below  and  to  the  right  of  the  same,  and  with  same  radius,  describe  a  circular  arc 
through  the  outside  or  inside  of  the  former  circle,  according  to  the  desired  position  of  the  shaded 
line,  as  shown  in  the  figure.  All  circles  and  circular  arcs  bounding  the  outside  of  a  surface  are 
shaded  according  to  circle  1.  All  those  bounding  inside  surfaces  are  shaded  according  to  circle 
2.  In  inking  Fig.  24,  observe  all  directions  for  inking  figures  22  and  23. 

Circles  and  circular  arcs  having  radii  less  than  \  in.  should  be  inked  by  means  of  the  bow 
compasses.  Their  use  is  similar  to  the  large  compasses,  save  in  shading,  when  it  is  not  neces- 
sary to  remove  the  needle  point  from  centre,  but  by  slightly  springing  the  same,  the  varying 
width  of  line  may  be  obtained. 


20  GEOMETRY. 

\ 

IT. 

GEOMETRICAL,    DEFINITIONS    AND    USEFUL    PROPOSITIONS. 

A  point  is  used  for  marking  positions  only,  and  has  no  dimension. 
A  line  is  produced  by  the  motion  of  a  point,  and  has  therefore  length  only. 
A  surface  is  produced,  by  the  motion  of  a  line,  and  has  length  and  breadth. 
A  solid  is  produced  by  the  motion  of  a  surface,  and  has  therefore   length,  breadth,  and 
thickness. 

L,  I  N  E  S. 

Lines  are  either  straight  or  curved.     They  may  be  horizontal,  vertical,  or  oblique. 
Parallel  lines  are  everywhere  equally  distant. 

ANGLES. 


\ 


RIGHT  ANGLE.  ACUTE  ANGLE.  OBTUSE  ANGLE 


When  two  straight  lines  meet,  they  form  an  angle.  There  are  three  kinds  of  angles,  Right, 
Acute,  and  Obtuse  angles. 

When  one  straight  line  meets  another  straight  line,  so  that  the  angles  on  either  side  are  equal, 
they  are  perpendicular  to  each  other  and  the  angles  are  Right  angles. 

An  Acute  angle  is  less  than  a  right  angle. 


GEOMETRY. 


21 


An  Obtuse  angle  is  greater  than  a  right  angle. 

When  one  straight  line  crosses  another  straight  line,  the  sum  of  the  four  angles  is  equal 
to  four  right  angles  ;  any  two  adjacent  angles  are  equal  to  two  right  angles,  and  are  said  to  be 
supplementary  to  each  other.  AED-|-DEBis  equal  to  two  right  angles,  and  D  E  B  is  the 
supplement  of  A  E  D. 

TRIANGLES. 


EQUILATERAL  TRIANGLE. 


ISOSCELES  TRIANGLE. 


SCALENE  TRIANGLE. 


A  triangle  is  a  figure  enclosed  by  three  straight  lines.  It  has  three  sides  and  three  angles. 
There  are  three  kinds  of  triangles,  named  after  their  sides,  viz  :  —  Equilateral,  Isosceles,  and 
Scalene. 

An  Equilateral  triangle  has  three  of  its  sides  equal. 

An  Isosceles  triangle  has  two  of  its  sides  equal. 

A  Scalene  triangle  has  none  of  its  sides  equal. 

There  are  three  kinds  of  triangles,  named  after  their  angles,  viz :  —  Right-angled  triangle, 
Acute-angled  triangle,  and  Obtuse-angled  triangle. 


EIGHT-ANGLED  TRIANGLE. 


ACUTE-ANGLED  TRIANGLE. 


OBTUSE-ANGLED  TRIANGLE. 


A  Right-angled  triangle  has  one  right  angle. 
An  Acute-angled  triangle  has  all  its  angles  acute. 


22  GEOMETRY. 

An  Obtuse-angled  triangle  has  one  of  its  angles  an  obtuse  angle. 

The  base  of  a  triangle  is  its  lowest  side.  The  vertex  of  a  triangle  is  the  angle  opposite 
the  base. 

The  perpendicular  height  or  altitude  of  a  triangle  is  measured  by  a  perpendicular  line  let  fall 
from  the  vertex  upon  the  base. 

The  sum  of  the  three  angles  of  any  triangle  is  equal  to  two  right  angles,  or  180°. 


QUADRILATERALS. 


SQUARE. 


RECTANGLE. 


RHOMBUS. 


RHOMBOID. 


A  figure  of  four  sides  is  called  a  Quadrilateral.  If  the  opposite  sides  are  equal  and  parallel, 
it  is  a  Parallelogram. 

There  are  four  kinds  of  parallelograms  :  — 

The  Square,  which  has  its  four  sides  equal,  and  all  of  its  angles  right  angles. 

The  Rectangle,  whose  opposite  sides  only  are  equal,  and  all  its  angles  are  right  angles. 

The  Rhombus,  whose  four  sides  are  equal,  and  none  of  whose  angles  are  right  angles. 

The  Rhomboid,  whose  opposite  sides  only  are  equal,  and  none  of  whose  angles  are  right 
angles. 

The  line  that  joins  any  two  opposite  angles  of  a  quadrilateral  is  called  its  diagonal. 


GEOMETRY. 
POLYGONS. 


23 


A  plane  figure  bounded  by  straight  lines  is  called  a  polygon.     The  term  is  usually  applied 


to  designate  figures  having  more  than  four  sides. 
The  Pentagon  has  five  sides. 
The  Hexagon  has  six  sides. 
The  Heptagon  has  seven  sides. 


The  Octagon  has  eight  sides. 
The  Nonagon  has  nine  sides. 
The  Decagon  has  ten  sides. 


CIRCLES. 


60 


7S 


Fiat.  2. 


Fio.  3. 


A  Circle  is  a  plane  figure  bounded  by  a  curved  line,  called  its  circumference,  all  points  of 
which  are  equally  distant  from  a  point  called  its  centre. 

A  Radius  is  a  straight  line  drawn  from  the  centre  to  the  circumference,  as  C  F  (Fig.  1). 
All  radii  of  the  same  circle  are  equal. 


24  GEOMETRY. 

The  Diameter  is  a  straight  line  drawn  through  the  centre,  terminating  in  the  circumference, 
as  A  B. 

An  Arc  of  a  circle  is  any  portion  of  its  circumference,  as  D  G  E  or  B  E.  An  arc  equal  to 
one-half  of  the  circumference  is  called  a  semi-circumference,  as  A  F  B. 

A  Chord  is  a  straight  line  which  has  its  extremities  in  the  circumference,  and  joins  the  ends 
of  an  arc,  as  D  E. 

A  Tangent  is  a  straight  line  which  touches  the  circumference,  but  does  not  intersect  it,  as 
F  H.  It  is  perpendicular  to  the  radius  at  the  point  of  tangency. 

A  Segment  of  a  circle  is  a  portion  of  a  circle  bounded  by  an  arc  and  a  chord,  as  the  area 
D  E  G.  A  segment  equal  to  one-half  of  the  circle  is  called  a  semi-circle,  as  the  area  A  B  E  G  D. 

A  Quadrant  is  the  fourth  part  of  a  circle,  as  B  C  G. 

Every  circle  is  supposed  to  be  divided  into  360  equal  parts,  called  degrees,  which  are  used  as 
a  measure  of  angles ;  therefore  the  arc  of  a  semi-circle  is  equal  to  180°.  The  arc  of  a  quadrant 
is  equal  to  90°. 

In  Fig.  3,  all  angles  indicated  by  the  full  lines  may  be  drawn  by  a  60°  triangle ;  all  those  in- 
dicated by  the  dotted  lines  may  be  drawn  by  a  45°  triangle  ;  those  indicated  by  the  broken  lines 
may  be  drawn  by  the  45°  and  60°  triangle,  as  shown  in  Figs.  7  and  8,  PI.  1. 

If  from  any  point  within  a  semi-circle  lines  be  drawn  to  the  extremities  of  the  diameter,  the 
included  angle  will  be  a  right  angle.  In  Fig.  2,  A  C  B,  A  D  B,  and  A  E  B  are  right  angles. 

SOLIDS. 

A  Polyhedron  is  a  solid  bounded  by  planes.  The  polyhedron  is  named  according  to  the  num- 
ber of  these  planes,  or  faces.  One  of  four  faces  is  called  a  tetrahedron,  one  of  six  faces  a  hexa- 
hedron, one  of  eight  faces  an  octahedron,  etc.  They  are  also  classified  according  to  the  shape 

and  relation  of  these  faces  as  follows :  — 

• 


GEOMETRY. 
PYRAMIDS. 


25 


REGULAR  OCTAHEDRON. 


REGULAR  PYRAMID. 


FRUSTUM  OP  PYRAMID. 


A  pyramid  is  a  polyhedron  one  face  of  which  is  a  polygon  and  known  as  the  base.  The  other 
faces  are  triangles  having  a  common  vertex  known  as  the  vertex  of  the  pyramid. 

The  altitude  of  a  pyramid  is  the  perpendicular  distance  from  the  vertex  to  the  base. 

A  pyramid  is  regular  if  its  base  is  a  regular  polygon  whose  centre  lies  in  the  perpendicular 
drawn  from  the  vertex  to  the  base. 

The  frustum  of  a  pyramid  is  that  pQrtion  of  a  pyramid  included  between  its  base  and  a  plane 
parallel  to  the  base. 

According  to  the  character  of  their  bases  pyramids  are  called  triangular,  rectangular,  pen- 
tagonal, etc. 

PRISMS. 


TRUNCATED  PRISM.  RIGHT  PRISM.  PARALLELOPIPED.  RIGHT  PARALLELOPIPED. 

A  prism  is  a  polyhedron  of  which  two  opposite  faces,  called  bases,  are  equal  and  parallel  poly- 
gons, and  the  other  faces,  called  lateral  faces,  are  parallelograms. 


26  GEOMETRY. 

The  altitude  is  the  perpendicular  distance  between  the  bases. 

A  truncated  prism  is  the  part  of  a  prism  included  between  the  base  and  a  cutting  plane  in- 
clined to  the  base. 

A  right  prism  is  one  whose  lateral  edges  are  perpendicular  to  the  bases. 

A  regularjpri&m  is  a  right  prism  whose  bases  are  regular  polygons. 

According  to  the  character  of  their  bases,  prisms  are  called  triangular,  quadrangular,  etc. 

A  prism  whose  bases  are  parallelograms  is  called  a  parallelepiped.  If  its  lateral  edges  are 
perpendicular  to  the  bases,  it  is  called  a  right  parallelepiped.  If  its  six  faces  are  all  rectangles, 
it  is  called  a  rectangular  parallelepiped.  If  its  six  faces  are  squares,  it  is  called  a  cube. 

CYLINDERS. 

A  cylindrical  surface  is  a  curved  surface  generated  by  the  motion  of  a  straight  line  which 
touches  a  given  curve  and  continues  parallel  to  itself. 

A  cylinder  is  a  solid  bounded  by  a  cylindrical  surface  and  two  parallel  planes  intersecting  this 
surface.  The  parallel  faces  are  called  bases. 

The  altitude  of  a  cylinder  is  the  perpendicular  distance  between  the  bases. 

A  circular  cylinder  is  a  cylinder  whose  base  is  a  circle. 

A  cylinder  of  revolution  is  a  cylinder  generated  by  the  revolution  of  a  rectangle  about  one 
side  as  an  axis. 

CONES. 

A  conical  surface  is  a  curved  surface  generated  by  the  motion  of  a  straight  line  one  end  of 
which  is  fixed  while  the  other  is  constrained  to  move  in  a  curve. 

A  cone  is  a  solid  bounded  by  a  conical  surface  and  a  cutting  plane  called  the  base. 

The  altitude  of  the  cone  is  the  perpendicular  distance  between  the  vertex,  or  fixed  point,  and 
the  base. 


GEOMETRICAL   PROBLEMS.  27 

A  circular  cone  is  a  cone  whose  base  is  a  circle. 

A  right  circular  cone  is  a  circular  cone  whose  axis  is  perpendicular  to  its  base.  It  is  also 
called  a  cone  of  revolution. 

SPHERE. 

A  sphere  is  a  solid  generated  by  the  revolution  of  a  semicircle  about  its  diameter. 

G-EOMETRICAL    PROBLEMS. 

GENERAL    INSTRUCTION. 

Divide  the  space  within  the  margin  line  into  twelve  rectangles,  and  place  one  example  in 
each  rectangle.  All  work  is  to  be  performed  with  HHHH  lead  pencil.  The  greatest  possible 
precision  must  be  used.  Construction  lines  are  to  be  made  very  fine  ;  given  and  required  lines 
being  made  stronger  by  pencilling  a  second  time.  Where  possible,  avoid  drawing  the  whole 
of  a  construction  line,  using  only  that  -portion  of  .the  line  necessary.  On  the  plates,  all  given 
and  required  lines  are  shown  in  full,  and  construction  lines  are  dotted.  When  two  methods  are 
given,  the  problem  will  be  constructed  by  the  practical,  or  draughtsman's  method,  and  tested  by 
the  geometrical. 

The  problems  are  not  to  be  copied  from  the  plates,  but  constructed  from  the  data  given  in  the 
examples  on  page  39.  In  doing  this  work  it  is  advisable,  first,  to  thoroughly  master  the  prob- 
lems relating  to  the  drawing  of  perpendiculars,  and  then  perform  the  examples  involving  these 
principles.  Next,  study  those  relating  to  angles,  and  similarly  perform  the  dependent  examples. 
Thus  continue  the  subject  according  to  the  divisions  indicated.  Place  the  number  of  the 
example  in  the  right  hand  lower  corner  of  the  square  containing  it. 


28  GEOMETRICAL    PROBLEMS. 

PERPENDICULARS. 

PLATE  4. 

Fig.  25.     PROBLEM  1.  —  To  bisect  a  straight  line,  A  B. 

With  centres  A  and  B,  and  any  radius  greater  than  one-half  of  A  B,  describe  arcs  1  and  2. 
Through  the  points  of  intersection  of  these  arcs  draw  a  line.  Its  intersection  with  the  line 
A  B,  at  C,  will  be  the  required  point.  The  more  practical  method  is  to  use  dividers  as  explained 
on  page  13. 

Fig.  26.     PROBLEM  2.  — To  bisect  an  arc  of  a  circle. 

Proceed  as  in  the  preceding  problem,  the  same  lettering  being  used. 

PROBLEM  3.  —  To  draw  a  perpendicular  to  a  straight  line. 

Practical  Method.  —  In  all  cases  where  the  given  line  has  been  drawn  by  the  T  square,  the 
perpendicular  may  be  drawn  by  sliding  the  triangle  along  the  edge  of  the  square  until  it  nearly 
coincides  with  the  given  point,  when  the  required  line  may  be  drawn.  If  the  given  line  makes 
an  angle  with  the  T  square,  the  construction  shown  in  Fig.  31  may  be  used. 

Fig.  27.     CASE  1.  —  When  the  point  is  on  the  line,  and  at,  or  near,  the  middle  of  the  line. 

Let  A  B  be  the  line,  and  C  the  point.  From  C,  with  any  radius,  draw  arcs  1,  and  from  the 
point  of  intersection  of  these  arcs  with  A  B,  with  any  radius  greater  than  arcs  1,  draw  arcs  2 
and  3.  The  line  drawn  through  the  point  of  intersection  of  these  arcs,  and  the  given  point  C, 
will  be  the  required  line. 

Fig.  28.  CASE  2.  —  When  the  point  is  on  the  line,  and  at,  or  near,  the  extremity  of  the 
line. 

First  Method.  —  Let  A  B  be  the  given  line,  and  A  the  given  point.  From  A,  with  any  radius, 
describe  arc  1.  With  centre  C,  arid  same  radius,  describe  arc  2.  Through  C,  and  the  point  of 
intersection  of  arcs  1  and  2,  draw  C  E,  and  with  same  radius  as  before,  from  intersection  of  arcs 


GEOMETRICAL    PROBLEMS.  29 

1  and  2,  describe  arc  3.  A  line  drawn  through  the  given  point  A,  and  the  point  of  intersection 
of  arc  3  and  line  C  E,  will  be  the  required  perpendicular. 

Second  Method.  —  From  B,  with  any  radius,  describe  arc  4.  From  point  D,  with  same  radius, 
describe  arc  5.  From  the  intersection  of  arcs  4  and  5  describe  arc  6.  From  the  intersection  of 
arcs  4  and  6  describe  arc  7.  The  line  drawn  through  this  last  point  of  intersection  F,  and  the 
given  point  B,  will  be  the  required  perpendicular. 

Fig.  29.  CASE  3.  —  When  the  point  is  outside  of,  and  opposite,  or  nearly  opposite,  the  middle 
of  the  line. 

Let  A  B  be  the  given  line,  C  the  given  point.  From  C,  with  as  great  a  radius  as  possible,  des- 
cribe arcs  1.  From  the  point  of  intersection  of  these  arcs  with  A  B,  with  same  radius,  describe 
arcs  2  and  3.  A  line  drawn  through  this  point  of  intersection,  and  the  given  point  C,  will  be  the 
required  line. 

Fig.  30.     CASE  4.  —  When  the  point  is  outside  of,  and  at,  or  near,  the  extremity  of  the  line. 

Let  A  B  be  the  given  line,  and  C  the  given  point.  From  C  draw  any  line  C  D.  Find  E,  the 
centre  of  C  D,  by  dividers  or  Problem  1.  On  C  D  as  a  diameter  describe  a  semi-circle.  Through 
the  given  point  C  and  the  intersection  of  semi-circle  with  A  B,  draw  C  F,  which  will  be  the 
required  perpendicular. 

ANGLES. 

PLATE  4. 

Fig.  32.     PROBLEM  4.  —  To  bisect  an  angle  A  B  C. 

From  B,  with  any  radius,  describe  arc  1.  From  its  points  of  intersection  with  A  B  and  C  D 
describe  arcs  2  and  3.  The  line  drawn  through  this  point  of  intersection,  and  B,  will  bisect  the 
given  angle. 


30  GEOMETRICAL    PROBLEMS. 

Fig.  33.     PROBLEM  5.  —  To  construct  an  angle  F  D  E  equal  to  a  given  angle  ABC. 

Draw  D  E.  From  B  and  D,  with  the  same  radius,  describe  arcs  1  and  2.  From  E,  with  radius 
equal  to  chord  A  C,  describe  arc  3.  Through  D,  and  point  of  intersection  of  arcs  2  and  3,  draw 
D  F,  the  remaining  side  of  the  required  triangle. 

Fig.  34.     PROBLEM  6.  —  To  construct  an  angle  of  45°  with  A  B  at  point  A. 

Practical.  —  With  T  square  draw  A  B,  and  with  45°  triangle  on  T  square  draw  A  D. 

Geometrical.  —  Through  the  given  point  A  describe  a  semi-circle  on  A  B  ;  draw  a  perpendic- 
ular through  the  centre  C.  A  line  drawn  through  the  point  A,  and  intersection  of  perpendicular 
with  semi-circle,  will  make  the  required  angle  with  A  B. 

Fig.  35.     PROBLEM  7.  —  To  construct  angles  of  30°  and  60°  with  A  B  at  point  A. 

Practical.  —  With  T  square  draw  A  B,  and  with  60°  triangle  on  T  square  draw  A  D  and  A  E. 

Geometrical.  —  Through  the  given  point  A  describe  a  semi-circle  on  A  B.  With  same  radius, 
from  C  describe  arc  1.  The  line  drawn  through  A,  and  this  point  of  intersection,  will  make  an 
angle  of  30°  with  A  B. 

From  the  given  point  A  as  a  centre,  with  any  radius,  describe  arc  2.  From  B,  with  the  same 
radius,  describe  arc  3.  A  line  drawn  through  A,  and  this  point  of  intersection,  will  make  an  angle 
of  60°  with  A  B. 

Fig.  36.     PROBLEM  8.  —  To  construct  an  angle  of  15°  with  A  B  at  point  A. 

Practical.  —  With  T  square  draw  A  B,  and  with  45°  and  60°  triangle  on  T  square,  as  in  Fig. 
7,  PI.  1,  draw  AC.  In  a  similar  manner  an  angle  of  75°  may  be  drawn.  See  dotted  position  in 
Fig.  7.  See  also  Fig.  8. 

Geometrical.  —  Construct  an  angle  of  30°  and  bisect  the  same,  as  A  C.  It  is  not  necessary 
to  draw  the  30°  line. 


GEOMETRICAL    PROBLEMS.  31 

TRIANGLES. 

PLATE  5. 

Fig.  37.     PROBLEM  9.  —  To  construct  an  equilateral  triangle,  having  given  the  side  A  B. 

Since  the  sides  are  equal,  the  angles  will  be  equal,  and  therefore  equal  to  60°,  since  their  sum 
is  equal  to  180°.  (See  page  22.) 

Practical.  —  Through  A  and  B,  with  60°  triangle,  draw  A  C  and  B  C. 

Geometrical.  —  With  A  and  B  as  centres,  and  radius  A  B,  describe  arcs  1  and  2.  From  point 
of  intersection  C,  draw  A  C  and  B  C. 

Fig.  38.  PROBLEM  10.  —  To  construct  an  isosceles  triangle,  having  given  the  base  A  B,  and 
either  the  equal  angles  C  A  B  and  C  B  A,  or  equal  sides  A  C  and  B  C. 

If  the  angles  be  given,  construct  CAB  and  C  B  A  equal  to  the  given  angle,  and  draw  A  C 
and  B  C  (See  Prob.  5). 

If  the  sides  be  given,  from  centres  A  and  B  draw  arcs  1  and  2  with  radius  equal  to  given 
sides.  From  point  of  intersection  C  draw  A  C  and  B  C. 

Fig.  39.  PROBLEM  11.  —  To  construct  a  scalene  triangle,  having  given  the  sides  A  B,  A  C, 
and  B  C. 

Draw  A  B.  With  centres  A  and  B,  and  radii  equal  to  given  sides,  draw  arcs  2  and  1.  Draw 
A  C  and  C  B. 

Fig.  40.  PROBLEM  12. —  To  construct  a  right-angled  triangle,  having  side  A  B  and  adjacent 
angle  C  A  B  given. 

Draw  B  C  perpendicular  to  A  B  ;  make  angle  CAB  equal  to  required  angle. 


32  GEOMETRICAL    PROBLEMS.    - 

TANGENTS. 
PLATE  5. 

PROBLEM  13.  —  To  draw  a  tangent  to  a  circle. 

Fig.  41.     CASE  1.  —  When  the  given  point  A  is  on  the  circle. 

Practical.  —  Place  the  triangle  to  coincide  with  centre  C  and  given  point  A,  as  though  to 
draw  A  C.  By  means  of  a  second  triangle  used  as  a  base,  turn  the  first  triangle  into  the  position 
shown  in  dotted  lines,  and  draw  A  B  perpendicular  to  A  C.  A  B  is  the  required  tangent. 

G-eometrical.  —  Draw  A  C  and  erect  a  perpendicular  at  A.     (Prob.  3,  Case  2.) 

Fig.  42.     CASE  2.  —  When  the  point  is  on  an  arc  of  the  circle,  and  the  centre  not  accessible. 

Practical.  —  From  the  given  point  A,  with  any  radius,  describe  arcs  1.  Place  the  edge  of 
triangle  to  coincide  with  points  B  and  D.  Draw  a  parallel  line  through  A.  This  will  be  the 
required  tangent. 

Fig.  43.  CASE  3.  —  W hen  the  given  point  B  is  without  the  circle.  In  this  case  two  tangents 
may  be  drawn. 

Practical.  —  From  the  given  point  B  draw  B  A  touching  the  circle.  Through  centre  C  draw 
perpendicular  to  A  B.  A  will  be  the  point  of  tangency.  In  like  manner  obtain  B  D. 

Geometrical.  —  On  B  C  as  a  diameter  describe  arc  1,  its  intersection  with  the  circle  at  A  and 
D  will  be  the  points  of  tangency.  The  angle  BAG,  inscribed  in  the  serni-circle,  will  be  a  right- 
angle.  (See  page  24.) 

Fig.  44.  PROBLEM  14.  —  To  draiv  a  circle  through  a  given  point  A,  and  tangent  to  given 
lines,  A  B  and  B  D. 

Since  the  circle  is  to  be  tangent  to  A  B  and  B  D,  its  centre  must  lie  upon  the  bisector  of  the 
angle  DBA.  Bisect  the  angle  DBA,  and  through  the  point  A  draw  A  C  perpendicular  to  A  B. 


GEOMETRICAL    PROBLEMS.  33 

C  will  be  the  centre  of  the  circle,  and  A  C  its  radius.  The  perpendicular  may  be  obtained  prac- 
tically by  triangles,  or  geometrically  as  shown. 

Fig.  45.  PROBLEM  15.  —  To  draw  any  number  of  circles  tangent  to  each  other  and  to  two 
given  lines  A  B  and  A  D. 

Bisect  angle  DAB,  and  with  any  radius,  H  K,  draw  a  circle  tangent  to  A  B  and  A  D.  From 
K  draw  K  E  perpendicular  to  A  K,  and  with  radius  E  K  describe  arc  4.  Through  F  draw  F  C 
perpendicular  to  A  B.  C  will  be  the  centre,  and  F  C  the  radius,  of  the  second  circle.  Repeat  the 
process. 

Fig.  46.     PROBLEM  16.  —  To  draw  a  tangent  to  two  given  circles  A  and  C. 

Join  A  and  C.  From  D  lay  off  D  H  equal  to  A  F.  With  centre  C,  and  radius  C  H,  draw 
arc  1.  From  A  draw  tangent  to  this  arc.  (Prob.  13,  Case  3.)  Through  B,  the  point  of  tan- 
gency,  draw  C  E,  and  through  A  draw  A  F  parallel  to  C  E.  E  and  F  will  be  the  points  of 
tangency,  and  E  F  the  tangent. 

Fig.  47.  PROBLEM  17.  —  To  draw  a  circle  of  a  given  radius  M,  tangent  to  two  given  circles 
B  and  C. 

Draw  indefinitely  in  any  direction,  lines  C  E  and  B  F.  Lay  off  H  E  and  K  F  equal  to  given 
radius  R,  and  through  E  and  F  describe  arcs  1  and  2  intersecting  at  A,  the  required  centre. 
These  arcs  will  also  intersect  in  a  second  centre. 

Fig.  48.  PROBLEM  18.  —  Through  three  given  points  A,  B,  and  D,  not  in  the  same  straight 
line,  to  draw  a  circle. 

Bisect  the  imaginary  chords  A  B  and  B  D.  The  point  of  intersection  C,  of  the  bisecting  lines, 
will  be  the  required  centre. 


34  GEOMETRICAL    PROBLEMS. 

INSCRIBED    AND    CIRCUMSCRIBED    POLYGONS. 

PLATE  6. 

Fig.  49.     PROBLEM  19.  —  To  circumscribe  a  circle  about  a  given  triangle  ABO. 

Bisect  two  of  the  sides,  as  A  C  and  B  C.  The  point  of  intersection  of  these  lines  will  be  the 
centre  of  the  required  circle.  Draw  circle  through  A,  B,  and  C.  (See  Prob.  18.) 

Fig.  50.     PROBLEM  20.  —  To  inscribe  a  circle  within  a  given  triangle  ABC. 

Bisect  two  of  the  angles,  as  C  A  B  and  ABC.  The  point  of  intersection  of  these  lines,  D, 
will  be  the  centre  of  the  required  circle.  From  this  point  draw  a  circle  tangent  to  lines  A  C,  C  B, 
and  A  B. 

Fig.  51.     PROBLEM  21.  —  To  inscribe  an  equilateral  triangle  within  a  circle. 

Practical.  —  Through  centre  D,  with  60°  triangle,  draw  A  D,  and  with  same  triangle  draw 
A  C  and  C  B.  Draw  A  B. 

G-eometrical.  —  From  any  point  in  the  circle,  as  C,  draw  arc  1  with  a  radius  equal  to  that  of 
the  circle.  From  its  intersection  with  the  circle,  and  with  the  same  radius,  draw  arc  2.  With 
chord  C  B,  and  centre  C,  describe  arc  3,  and  connect  points  A,  B,  and  C,  which  will  give  the 
required  triangle. 

Fig.  52.     PROBLEM  22.  —  To  inscribe  a  square  within  a  circle. 

Practical.  —  With  45°  triangle  draw  two  perpendicular  diameters,  A  C  and  B  D,  and  connect 
points  A,  B,  C,  and  D,  which  will  give  the  required  square. 

G-eometrical. — Draw  any  diameter  B  D.  Draw  a  second  diameter  A  C  perpendicular  to  it- 
Connect  points  A,  B,  C,  and  D  as  before. 

Fig.  53.     PROBLEM  23.  —  To  inscribe  a  pentagon  within  a  circle. 

Draw  any  diameter  G  F,  and  a  radius  A  K  perpendicular  to  it.     Bisect  K  F,  and  with  H  as  a 


GEOMETRICAL    PROBLEMS.  35 

centre,  and  radius  A  H,  describe  arc  8.  With  centre  A,  and  radius  A  L,  describe  arc  4.  A  B  is 
the  side  of  a  pentagon.  Obtain  the  remaining  points  by  describing  arcs  5,  6,  and  7  with  same 
radius.  Connect  points  A,  B,  C,  D,  and  E  to  obtain  the  required  pentagon. 

CONSTRUCTION    OF    THE    HEXAGON. 

PLATE  6. 

Fig.  54.     PROBLEM  24. —  To  inscribe  a  Hexagon  within  a  circle. 

Practical.  —  Draw  a  horizontal  diameter  P  C.  With  60°  triangle  draw  diameter  E  B.  Draw 
A  B  and  E  D,  and  with  triangle  draw  B  C,  F  E,  A  F,  and  C  D. 

Geometrical.  —  Draw  diameter  F  C.  With  centres  F  and  C,  and  radius  equal  to  that  of  the 
circle,  draw  arcs  1  and  2.  Connect  points  of  intersection  A,  B,  C,  D,  E,  and  F,  to  obtain  the 
required  hexagon. 

Observe  that  the  angles  at  the  centre,  as  B  K  C,  are  60°.     (See  Prob.  7.) 

Fig.  55.     PROBLEM  25.  —  To  circumscribe  a  hexagon  about  a  circle. 

Practical.  —  With  60°  triangle  draw  diameters  A  D  and  E  B,  and  with  same  triangle  draw 
sides  A  B  and  E  D,  E  F  and  B  C,  A  F  and  C  D,  each  tangent  to  the  given  circle. 

Geometrical.  —  Draw  any  diameter  A  D.  With  H  as  centre,  and  radius  equal  to  that  of  the 
circle,  describe  arc  1.  Bisect  the  arc  H  L  N,  and  through  L  draw  A  B  parallel  to  H  N.  With 
centre  K,  and  radius  A  K,  describe  circle  ACE.  In  this  inscribe  a  hexagon  by  Prob.  24. 

Fig.  56.     PROBLEM  26.  —  To  draw  a  hexagon,  having  given  a  long  diameter  A  D. 

Practical.  —  With  dividers  find  centre  K.  With  60°  triangle,  through  A  and  D,  draw  A  B 
and  D  E.  With  same  triangle,  through  K  draw  B  E,  and  through  D  and  A  draw  C  D  and  A  F. 
Draw  B  C  and  F  E,  completing  the  required  hexagon.  The  point  B  may  be  obtained  without 


UNIVERSITY ; 

lie 


36  GEOMETRICAL    PROBLEMS. 

finding  K,  by  drawing  a  line  through  D  at  an  angle  of  30°  with  A  D.  From  its  intersection  with 
A  B,  B  C  may  be  drawn.  < 

Geometrical.  —  Bisect  A  D.  With  K  as  centre  describe  circle  A  C  E,  and  in  this  inscribe  a 
hexagon  by  Prob.  24. 

Fig.   57.     PROBLEM  27.  —  To  draw  a  hexagon,  having  given  the  short  diameter  G^H. 

Practical.  —  With  dividers  find  centre  K.  With  60°  triangle  draw  indefinitely  F  C  ;  also 
through  G  and  H  draw  perpendiculars  F  E  and  B  C.  Through  centre  K'draw  A  D,  and  with 
triangle  draw  the  remaining  sides  F  A  and  D  C,  A  B  and  D  E. 

Geometrical.  —  Bisect  G  EL  From  G  draw  perpendicular  G  F.  From  K  draw  F  K  C,  making 
an  angle  of  30°  with  G  K.  Through  point  of  intersection  F,  and  with  centre  K,  draw  circle  F  B  D, 
and  inscribe  a  hexagon  by  Prob.  24. 

Fig.  58.     PROBLEM  23.  —  To  draw  a  hexagon,  having  given  a  side  A  B. 

Practical.  — Through  A  and  B,  with  60°  triangle,  draw  indefinitely  A  D  and  B  B,  and  through 
their  intersection  K,  draw  F  C.  Draw  F  A,  B  C,  F  E,  D  C,  and  E  D. 

Geometrical.  —  With  centres  A  and  B,  and  radius  A  B,  describe  arcs  1  and  2.  From  point 
of  intersection  K,  with  same  radius,  describe  circle  A  E  C.  Inscribe  a  hexagon  by  Prob.  24. 

NOTE.  —  Observe  that  the  above  problems  relating  to  the  hexagon  may  be  performed  by  use 
of  60°  triangle  and  T  square  only.  To  attain  the  ne'cessary  familiarity  with  these  problems  they 
should  be  performed  many  times. 

Fig.  59.     PROBLEM  29. —  To  inscribe  an  oc'agon  within  a  given  circle  A  C  E  G. 

Practical.  —  With  45°  triangle  and  J  square  draw  diameters  A  E,  G  C,  F  B,  and  H  D. 
Connecting  the  points  of  intersection  with  the  circle  will  give  the  inscribed  octagon. 

Geometrical.  —  Draw  any  diameter  G  C.  At  centre,  and  perpendicular  to  it,  draw  A  E. 
Bisect  A  K  G  and  A  K  C.  Connect  points  of  intersection  with  circle  as  before. 


GEOMETRICAL    PROBLEMS.  37 

Fig.  60.     PROBLEM  30. —  To  circumscribe  an  octagon  about  a  circle  A  B  C  D. 

Practical.  —  With  45°  triangle  and  T  square  draw  tangents  F  E  and  L  N,  G  H  and  P  0. 
At  45°,  draw  also  tangents  F  G,  H  L,  etc.,  completing  the  octagon. 

G-eometrical.  —  Draw  the  perpendicular  diameters  A  C  and  B  D.  With  centres  A,  B,  C,  and 
D,  and  radius  A  K,  describe  arcs  1,  2,  3,  4.  By  connecting  these  points  of  intersection,  a  circum- 
scribed square  will  be  obtained.  With  the  centres  R,  S,  etc.,  and  radius  R  K,  describe  arcs  5,  6, 
etc.  to  obtain  the  points  G,  H,  L,  N,  etc.,  which,  being  connected,  will  complete  the  circumscribed 
octagon. 
•< 

INSCRIBED   AND    CIRCUMSCRIBED    CIRCLES. 

PLATE  7. 

Fig.  61.  PROBLEM  31. — Within  an  equilateral  triangle  ABC  to  draw  three  equal  circles 
tangent  to  each  other  and  one  side  of  the  triangle. 

Bisect  the  angles  A,  B,  and  C.  Bisect  the  angle  D  C  A.  E  is  the  centre  of  one  of  the  required 
circles.  With  centre  K,  and  radius  K  E,  describe  arc  E  F  G.  F  and  G  will  be  the  remaining 
centres.  From  these  centres,  with  E  L  as  radius,  describe  the  required  circles. 

Fig.  62.  PROBLEM  32.^  Within  an  equilateral  triangle  A  B  C  to  draw  three  equal  circles 
tangent  to  each  other  and  two  sides  of  the  triangle, 

Bisect  angles  A,  B,  and  C.  Bisect  angle  D  C  A.  E  will  be  the  centre  of  one  of  the  circles. 
With  K  as  centre,  and  radius  K  E,  describe  arc  E  F  G  to  obtain  the  remaining  centres.  Draw 
circles  tangent  to  sides  A  B,  B  C,  and  A  C. 

NOTE.  —  Instead  of  bisecting  the  angle  D  C  A,  to  obtain  one  of  the  centres,  describe  a  semi- 
circumference  A  F  C  on  one  of  the  sides,  as  A  C,  thus  obtaining  centre  F. 


38  GEOMETRICAL    PROBLEMS. 

Fig.  63.  PROBLEM  33.  —  Within  an  equilateral  triangle  A  B  C  to  draw  six  equal  circles 
which  shall  be  tangent  to  each  other  and  the  sides  of  the  triangle. 

Bisect  angles  and  obtain  E  as  in  Prob.  31.  Through  E  draw  H  N  parallel  to  A  C.  Draw 
H  M  parallel  to  A  B,  and  M  N  parallel  to  B  C.  With  E,  H,  F,  M,  G,  and  N  as  centres,  and  with 
radius  E  L,  describe  the  required  circles. 

Fig.  64.  PROBLEM  34.  —  Within  a  given  circle  A  C  E  to  draw  three  equal  circles  tangent 
to  each  other  and  the  given  circle. 

Divide  the  circle  into  six  equal  parts,  by  diameters  A  D,  B  E,  C  F.  Produce  A  D  indefinitely, 
and  from  E  draw  tangent  E  G.  Bisect  D  G  E.  With  K  as  centre,  and  radius  H  K,  describe  arc 
H  L  M,  and  with  radius  H  E,  from  centres  H,  L,  and  M,  describe  the  required  circles. 

Fig.  65.  PROBLEM  35.  —  Within  a  given  circle  to  draw  any  number  of  equal  circles  tangent 
to  each  other  and  the  given  circle. 

General  Method.  — Divide  the  circle  by  diameters  into  twice  as  many  equal  parts  as  circles 
required  ;  in  this  case,  eight.  Suppose  the  centre  of  one  of  these  circles  to  lie  on  A  K ;  then  the 
circle  must  be  tangent  to  both  F  K  and  K  B.  Draw  tangent  at  A  intersecting  K  B.  Since  the 
required  circle  must  be  tangent  to  the  given  circle  at  A,  it  will  also  be  tangent  to  A  B,  and  as 
it  must  now  lie  in  the  angles  F  K  B  and  A  B  K,  its  centre  must  lie  at  D,  the  intersection  of 
their  bisectors.  With  centre  K  draw  circle  through  D,  and  its  intersection  with  E  K,  C  K,  etc., 
will  give  the  required  centres.  From  these  centres,  and  with  radius  A  D,  describe  the  required 
circles. 

Fig.  66.  PROBLEM  36.  —  About  a  given  circle  to  circumscribe  any  number  of  equal  circles 
tangent  to  each  other  and  the  given  circle. 

Divide  the  circle  by  diameters  into  twice  as  many  equal  parts  as  circles  required  ;  in  this  case, 
six.  From  extremity  A  of  any  diameter,  draw  tangent  A  B.  Produce  K  B,  making  B  C  equal 


GEOMETRICAL    EXAMPLES.  39 

to  A  B.  At  C,  perpendicular  to  B  C,  draw  C  D,  intersecting  A  K  produced  at  D.  This  will  be 
the  centre,  and  A  D  will  be  the  radius,  of  one  of  the  required  circles.  With  centre  K,  and  radius 
D  K,  obtain  other  centres,  and  describe  circles  as  in  preceding  problem. 

EXAMPLES. 
NOTE.  — For  Examples  1  to  40  inclusive,  divide  the  sheet  into  12  rectangles. 

u  «  4^    ..    ^g  4;  «  a  Q  *t 

PERPENDICULARS. 

1.  Bisect  a  line  3  in.  long. 

2.  Bisect  an  arc  of  2£  in.  radius  and  2|  in.  chord. 

3.  Draw  a  line  2|  in.  long,  and  erect  a  perpendicular  If  in.  from  one  end.     Do  not  use  T  square. 

4.  Draw  a  line  2^  in.  long,  and  erect  perpendiculars  at  the  extremities.     Use  both  methods. 

5.  From  a  point  nearly  over  the  centre  of  a  line  3|  in.  long  draw  a  perpendicular  to  the  line. 

6.  From  a  point  nearly  over  the  extremity  of  a  line  2 ^  in.  long,  draw  a  perpendicular  to  the 

line.     Do  not  use  T  square. 

ANGLES. 

7.  Draw  two  lines  intersecting  and  making  any  angle.     Construct  a  similar  angle,  and  bisect 

the  same. 

8.  From  one  extremity  of  a  line  3  in.  long  draw  a  second  line  making  an  angle  of  45°  with  the 

first.     Similarly  construct  an  angle  of  30°  at  the  other  end. 

9.  From  one  extremity  of  a  line  3  in.  long  draw  a  second  line  making  an  angle  of  22i°  with  the 

first.     Similarly  construct  an  anffle  of  15°  at  the  other  end. 


40  GEOMETRICAL    EXAMPLES. 

10.  From  one  extremity  of  a  line  3  in.  long  draw  a  second  line  making  an  angle  of  60°  with  the 

first.     Similarly  construct  an  angle  of  75°  at  the  other  end. 

11.  Describe  a  circle  3  in.  in  diameter,  and  divide  it  into  angles  of  15°  by  means  of  triangles  and 

T  square. 

TRIANGLES. 

12.  Draw  an  equilateral  triangle  having  a  base  2^1  in. 

13.  Construct  an  isosceles  triangle  having  a  base  2  in.  and  the  equal  sides  2f  in. 

14.  Construct  an  isosceles  triangle  having  a  base  of  1-|  in.  and  the  equal  angles  75°. 

15.  Construct  an  isosceles  triangle  having  a  base  of  3  in.  and  the  angle  at  the  vertex  150°. 

16.  Construct  a  scalene  triangle  having  sides  2J,  2|,  and  3  if  in. 

17.  Construct  a  right  angle  triangle  having  base  2|  in.  and  one  angle  of  30°. 

TANGENTS. 

18.  Draw  a  tangent  to  a  circle  2|  in.  in  diameter. 

19.  Draw  a  tangent  to  the  middle  point  of  the  arc  of  a  circle  of  2J  in.  radius,  having  a  chord 

of  2|  in.     Do  not  use  the  centre  of  circle. 

20.  Draw  a  tangent  to  a  circle  2|  in.  in  diameter  from  a  point  2|  in.  from  centre  of  circle. 

21.  Draw  two  lines  making  an  angle  of  45°  with  each  other,  and  a  circle  tangent  to  these  lines 

at  a  point  If  in.  from  vertex  of  the  angle. 

22.  Draw  two  lines  making  an  angle  of  30°  with  each  other,  and  two  circles  tangent  to  each 

other  and  these  lines.     The  diameter  of  the  smaller  circle  is  f  in. 

23.  Draw  a  tangent  to  two  circles  having  diameters  of  If  and  |  in.  and  their  centres  2  in. 

apart. 


GEOMETRICAL    EXAMPLES.  41 

24.  Draw  a  circle  having  a  diameter  of  1|  in.  tangent  to  two  circles  whose  diameters  are  If 

and  1£  in.  and  whose  centres  are  If  in.  apart. 

25.  Prescribe  three  points  and  draw  a  circle  through  them. 

INSCRIBED    AND    CIRCUMSCRIBED    POLYGONS. 

26.  Circumscribe  a  circle  about  a  scalene  triangle  having  sides  1|  in.,  2|  in.,  and  2f  in. 

27.  Inscribe  a  circle  within  an  isosceles  triangle  whose  base  is  2|  in.  and  equal  sides  3|  in. 

28.  Draw  a  right-angle  triangle  having  a  base  of  2^  in.  and  one  of  the  oblique  angles  30°.     Cir- 

cumscribe a  circle  about  this  triangle. 

29.  Within  a  circle  3  in.  in  diameter  inscribe  an  equilateral  triangle 

30.  Within  a  circle  3  in.  in  diameter  inscribe  a  square. 

31.  Within  a  circle  3  in.  in  diameter  inscribe  a  pentagon. 

32.  Within  a  circle  3  in.  in  diameter  inscribe  a  hexagon. 

33.  About  a  circle  2|  in.  in  diameter  circumscribe  a  hexagon. 

HEXAGONS. 

34.  Draw  a  hexagon  having  its  long  diameter  2|  in. 

35.  Draw  a  hexagon  having  a  short  diameter  2|  in. 

36.  Draw  a  hexagon  having  one  side  1|  in. 

37.  Draw  a  hexagon  having  one  side   1|-  in.  long  and  at  an  angle  of   45°   with  the  hori- 

zontal. 

38.  Draw  a  hexagon  having  its  short  diameter  2|  in.  and  one  side  horizontal. 

39.  Within  a  circle  3  in.  in  diameter  inscribe  an  octagon. 

40.  Circumscribe  an  octagon  about  a  circle  2£  in.  in  diameter 


42  GEOMETRICAL    EXAMPLES. 


INSCRIBED    AND    CIRCUMSCRIBED    CIRCLES. 

41.  Within  an  equilateral  triangle  having  sides  of  4  in.  draw  3  equal  circles  touching  each  other 

and  one  side  of  the  triangle. 

42.  Within  an  equilateral  triangle  having  sides  of  4  in.  draw  3  equal  circles  touching  each  other 

arid  two  sides  of  the  triangle. 

43.  Within  an  equilateral  triangle  having  sides  of  4  in.  draw  6  equal  circles  which  shall  be 

tangent  to  each  other  and  the  sides  of  the  triangle. 

44.  Within  a  circle  4  in.  in  diameter  draw  3  equal  circles  tangent  to  each  other  and  the  given 

circle. 

45.  Within  a  circle  4  in.  in  diameter  inscribe  5  equal  circles  tangent  to  each  other  and  the 

given  circle. 

46.  About  a  circle  If  in.  in  diameter  circumscribe  5  equal  circles  tangent  to  each  other  and 

the  given  circle. 


CONIC    SECTIONS. 


43 


III. 

CONIC    SECTIONS. 


If  a  cone  be  cut  by  three  planes,  in  the  manner  indicated  in  the  figures,  three  important  curves 
will  be  derived,  viz :  the  Ellipse,  Parabola,  and  the  Hyperbola. 

The  Ellipse  is  obtained  by  a  cutting  plane  oblique  to  the  axis,  and  making  a  greater  angle 
with  it  than  DAB,  Fig.  1.  It  is  a  closed  or  continuous  curve,  and  is  symmetrical  with  respect 
to  two  perpendicular  axes.  See  PI.  8. 

The  Parabola  is  obtained  by  a  cutting  plane  parallel  to  A  B,  or  making  a  smaller  angle  with 
the  axis  than  A  B,  Fig.  2.  It  is  not  a  closed  curve.  The  branches  of  the  curve  continually 
approach  parallelism  but  become  parallel  only  at  infinity.  See  PL  9,  Figs.  71  and  72. 

The  Hyperbola  is  obtained  by  a  cutting  plane  parallel  to  the  axis  of  the  cone.  Fig.  3.  In 
this  curve  the  branches  are  always  diverging.  See  PI.  9,  Figs.  73  and  74. 


44  CONIC    SECTIONS. 

THE    ELLIPSE. 

PLATE  8. 

The  Ellipse  is  a  curve  generated  by  a  point  moving  in  a  plane  so  that  the  sum  of  the  distances 
from  this  point  to  the  two  fixed  points  shall  ahvays  be  constant. 

The  fixed  points,  E  and  F,  Fig.  67,  are  called  the/ocz.  They  lie  on  the  longest  line  that  can 
be  drawn,  terminating  in  the  curve  of  the  ellipse.  The  line  is  known  as  the  major  axis,  and  the 
perpendicular  to  it  at  its  middle  point,  also  terminating  in  the  ellipse,  is  the  minor  axis.  Their 
intersection  is  called  the  centre  of  the  ellipse,  and  lines  drawn  through  this  point,  and  terminating 
in  the  ellipse,  are  known  as  diameters.  When  two  such  diameters  are  so  related  that  a  tangent 
to  the  ellipse  at  the  extremity  of  one  is  parallel  to  the  second,  they  are  called  conjugate  diameters. 
K  L  and  M  N  are  two  such  diameters. 

In  order  to  construct  an  ellipse,  it  is  in  general  necessary  that  either  of  the  following  be  given  : 
the  major  and  minor  axes ;  either  axis  and  the  foci ;  two  conjugate  diameters ;  a  chord  and  axis 
perpendicular  to  it. 

Fig.  67.  FIRST  METHOD.  —  By  definition  it  may  be  seen  that  a  series  of  points  must  be 
so  chosen  that  the  sum  of  the  distances  from  either  of  them  to  the  foci  must  equal  the  major 
axis.  Thus  H  E  +  H  F  must  equal  C  E  +  C  F  and  K  F  +  K  E,  each  being  equal  to  A  B.  If 
then  the  major  axis  and  foci  be  given  to  draw  the  curve,  points  may  be  determined  as  follows  : 
From  E,  with  any  radius  greater  than  A  E  and  less  than  E  B,  describe  an  arc.  From  F,  with  a 
radius  equal  to  the  difference  between  the  major  axis  and  the  first  radius,  describe  a  second  arc 
cutting  the  first.  The  points  of  intersection  of  these  arcs  will  be  points  the  sum  of  whose  dis- 
tances from  the  foci  will  equal  the  major  axis,  and  therefore  points  of  an  ellipse.  Similarly  find  as 
many  points  as  may  be  necessary  to  enable  the  curve  to  be  drawn  freehand.  Draw  very  lightly, 


CONIC   SECTIONS.  45 

using  a  sharp  point  and  describing  an  almost  invisible  curve.  Mechanical  methods  should  be 
used  for  inking  only.  -Do  not  erase  construction  lines. 

Having  given  the  major  and  minor  axes,  we  can  find  the  foci  by  describing  from  C  as  centre 
an  arc  with  a  radius  equal  to  one-half  the  major  axis  A  B.  The  points  of  intersection  with  the 
major  axis  will  be  the  foci ;  and  this  must  be  so,  since  the  sum  of  these  distances  is  equal  to  the 
major  axis  ;  and  the  point  C  being  mid-way  between  A  and  B,  the  two  lines  C  E  and  C  F  must 
be  equal.  Again,  if  we  have  the  major  axis  and  foci  given,  we  may,  with  a  radius  equal  to  one- 
half  this  axis,  describe  arcs  from  the  foci,  cutting  the  perpendicular  drawn  at  middle  point  of 
major  axis,  and  thus  obtain  the  minor  axis.  Having  the  two  axes,  proceed  as  before  to  describe 
the  ellipse. 

A  tangent  to  an  ellipse  may  be  drawn  at  any  point  K,  by  producing  K  F,  and  bisecting  the 
angle  S  K  E ;  the  bisecting  line  K  T  will  be  the  required  tangent. 

Do  not  copy  the  problems  from  the  plates,  but  construct  from  the  data  given  in  the  examples 
on  page  50. 

Fig.  68.     SECOND  METHOD.  —  To  describe  an  Ellipse  by  Trammels. 

Let  A  B  and  C  D  be  the  major  and  minor  axes  of  an  ellipse.  Lay  off  on  a  piece  of  paper, 
having  a  clean  cut  edge,  the  distance  R  T,  equal  to  one-half  the  major  axis,  and  R  S,  equal  to 
one-half  the  minor  axis.  Now,  if  point  T  be  placed  upon  the  minor  axis,  and  point  S  upon  the 
major  axis,  and  the  paper  constrained  to  move  always  under  these  conditions,  the  point  R  will 
describe  an  ellipse.  If  points  be  laid  off  upon  the  drawing  to  correspond  with  the  different 
positions  of  R,  a  number  of  points  may  be  obtained  through  which  the  required  ellipse  may  be 
drawn.  This  is  the  best  method  for  a  draughtsman,  since  construction  lines  are  not  required. 

Fig.  68.  THIRD  METHOD. —  To  describe  an  approximate  ellipse,,  the  major  and  minor  axes 
being  given. 


46  CONIC    SECTIONS. 

For  many  purposes  in  drawing  it  is  sufficiently  accurate  to  describe  the  ellipse  by  means  of 
four  circular  arcs  of  two  different  radii.  The  following  is  one  of  several  methods  employed  for 
describing  an  ellipse  from  four  centres. 

On  the  minor  axis  lay  off  from  the  centre  G  F  and  G  0  equal  to  the  difference  between  the 
major  and  minor  axes  (A  B  —  CD).  Next  lay  off  points  E  and  L  on  the  major  axis,  their  dis- 
tance from  the  centre  being  equal  to  three-quarters  of  G  F.  Connect  points  F,  E,  0,  and  L,  and 
produce  the  same.  With  centre  E,  and  radius  A  E,  describe  arc  L  A  H.  With  centre  F,  and  radius 
F.  D,  describe  arc  H  D  K.  In  similar  manner  describe  the  two  remaining  arcs,  K  B  R  and  R  C  L. 

Do  not  use  this  method  when  the  major  axis  is  twice  as  great  as  the  minor  axis. 

Figs.  69  and  70.  FOURTH  METHOD.  —  This  is  a  very  general  method,  and  may  be  used 
when  we  have  given,  either  the  major  and  minor  axes,  one  of  the  axes  and  a.  chord  of  the  ellipse, 
or  any  two  conjugate  diameters.  The  number  of  construction  lines  required,  make  this  method 
unsuitable  for  the  draughtsman. 

Fig.  69.     CASE  1.  —  Having  given  the  major  and  minor  axes,  to  describe  an  ellipse. 

Draw  B  6  parallel  to  the  minor  axis,  and  divide  into  any  number  of  equal  parts,  in  this  case, 
six.  Divide  B  G  into  the  same  number  of  equal  parts.  Through  points  1,  2,  3,  etc.,  on  B  6,  draw 
lines  to  extremity  C,  of  minor  axis.  From  the  otber  extremity  of  the  minor  axis  D  draw  lines 
through  points  1,  2,  3,  etc.,  on  B  G,  intersecting  the  above  lines  in  points  which  will  lie  in  the 
required  ellipse.  Construct  remainder  of  ellipse  in  the  same  manner. 

Fig.  69.     CASE  2.  —  Having  given  an  axis  C  Z),  and  chord  F  H. 

From  F  draw  F  4  parallel  to  C  D  ;  divide  it  into  any  number  of  equal  parts,  in  this  case,  four. 
Divide  the  half  chord  F  E  into  the  same  number  of  equal  parts  ;  through  these  points  and  extremi- 
ties of  given  axis  draw  intersecting  lines  as  before,  thereby  obtaining  the  elliptical  arc  F  D. 
Construct  opposite  side  in  the  same  manner. 


CONIC    SECTIONS.  47 

CASE  3.  —  Having  given  the  conjugate  diameters  A  B  and  C  D. 

From  A  and  B  draw  lines  A  6  and  B  6  parallel  to  the  diameter  C  D.  Divide  these  into  any 
number  of  equal  parts,  and,  having  divided  B  G  and  A  G  into  the  same  number  of  equal  parts, 
draw  lines  from  these  points  to  the  extremities  of  diameter  C  D,  similar  to  Case  1.  Iii  the  same 
manner  describe  opposite  side. 

THE    INKING    OF    CURVES. 

Use  the  following  method  for  the  inking  of  all  curves  other  than  circular  arcs.  Suppose  it  is 
required  to  ink  the  curve  G  A  F  H,  PI.  1,  Fig.  5.  Wherever  possible  use  the  compasses.  In  this 
case,  a  circular  arc  struck  from  the  centre  K  will  be  seen  to  coincide  with  the  required  curve, 
through  points  G,  A,  B,  and  C.  Just  at  the  point  C  it  will  appear  to  leave  the  curve,  and  should 
therefore  not  be  drawn  up  to  that  point,  but  the  inked  line  should  stop  at  the  point  B.  Since  the 
radius  required  to  complete  the  curve  would  now  be  too  large  for  the  compasses,  select  such  a  scroll 
as  will  coincide  with  the  largest  portion  of  the  curve,  beginning  at  the  point  A,  a  little  to  the  left  of 
the  line  already  inked.  In  the  figure,  this  curve  is  seen  to  coincide  with  the  line  up  to  the  point 
F,  but  the  inked  portion  of  this  line  should  stop  short  of  the  point  F,  as  at  E.  Now  another 
portion  of  this,  or  some  other  scroll,  should  be  made  to  coincide  with  the  remaining  portion  of 
the  required  lines,  beginning  at  least  as  far  back  as  the  point  D,  thus  laying  the  scroll  to  coincide 
with  a  greater  portion  of  the  curve  than  is  required  to  be  inked  at  any  one  stroke. 

In  inking  the  ellipse,  PI.  8,  Fig.  67,  X  B  Y  and  W  A  V  should  first  be  inked,  and  by  means 
of  the  compasses.  Next,  obtain,  if  possible,  a  scroll  to  coincide  with  arc  C  H  X,  and  for  a  short 
distance  on  either  side  of  points  C  and  X,  so  as  to  insure  a  perfect  copy  of  the  curve.  The  line 
should  now  be  drawn  from  the  point  C  to  the  point  X,  but  never  should  it  pass  to  the  left  of  the 
point  C,  with  scroll  in  this  position.  Now,  marking  upon  the  scroll  (which  should  be  made  of 


48  CONIC    SECTIONS. 

white  wood),  a  point  which  coincides  with  the  point  C,  and  reproducing  this  point  upon  the  opposite 
side  of  the  scroll,  reverse  the  scroll  in  order  to  draw  lines  to  the  left  of  C.  Similarly  ink  lower 
half  of  ellipse. 

The  greatest  possible  care  must  be  used  in  inking  these  curves,  as  no  class  of  line  work 
requires  more  skill  and  patience. 

Construction  lines  should  be  left  in  pencil. 

THE    PARABOLA. 

PLATE  9. 

The  Parabola  is  a  curve  generated  by  a  point  moving  in  a  plane  so  that  its  distance  from  a 
given  point  shall  be  constantly  equal  to  its  distance  from  a  given  right  line. 

The  given  point  F,  Fig.  71,  is  the  focus ;  the  given  right  line  C  D,  the  directrix.  The  line 
A  B,  perpendicular  to  C  D,  through  F,  is  the  axis,  and  V,  its  intersection  with  the  curve,  is  the 
vertex. 

Fig.  71  FIRST  METHOD.  —  Having  given  the  focus  F,  and  the  directrix  C  D,  to  describe  a 
parabola. 

Bisect  F  A  to  find  the  vertex  V.  Through  any  point  on  the  axis,  as  L,  draw  M  N  parallel 
to  the  directrix,  and  with  radius  L  A,  from  focus  F  as  centre,  describe  arcs  1,  intersecting  line 
M  N  at  points  M  and  N.  Obtain  other  points  in  a  similar  manner,  and  through  the  points  draw 
the  curve  freehand. 

A  tangent  to  the  curve  may  be  drawn  at  any  point,  as  M,  by  drawing  M  0  parallel  to  the  axis 
and  bisecting  angle  O  M  F.  M  T  is  a  tangent  to  the  curve  at  point  M. 

Fig.  72.     SECOND  METHOD.  —  Having  given  the  height  V  B,  and  the  breadth  G-  E. 


CONIC    SECTIONS.  49 

Draw  A  E  and  C  G  parallel  and  equal  to  V  B.  Divide  A  E  and  E  B  into  the  same  number 
of  equal  parts.  From  the  divisions  on  B  E  draw  parallels  to  the  axis,  and  from  the  divisions  on 
A  E  draw  lines  converging  to  the  vertex  V.  The  intersection  of  these  lines  1  and  1,  2  and  2, 
etc.,  will  give  points  in  the  required  curve.  Produce  opposite  side  in  the  same  manner. 

THE    HYPERBOLA. 

PLATE  9. 

The  Hyperbola  is  a  curve  generated  by  a  point  moving  in  a  plane,  so  that  the  difference  of  its 
'distances  from  tivo  fixed  points  shall  be  constantly  equal  to  a  given  line. 

The  two  fixed  points,  F  and  F',  Fig.  73,  are  the  foci.  The  given  line,  V  V,  is  the  trans- 
verse axis. 

Fig.  73.  FIRST  METHOD.  —  Having  given  the  axis  V  V' ',  and  the  foci  F  and  F',  to  describe 
an  hyperbola. 

With  any  radius,  from  centres  F  and  F',  describe  arcs  1  ;  from  same  centres,  with  radius 
increased  by  V  V,  describe  arcs  2  intersecting  the  arcs  1.  These  points  of  intersection  will  be 
points  in  the  required  curve. 

Fig.  74  SECOND  METHOD.  — Having  given  the  axis  V  V\  the  height  of  the  curve  V  £,  and 
its  breadth  A  D. 

Draw  A  C  and  D  E  parallel  to  B  V.  Divide  lines  A  B  and  A  C  into  any  number  of  equal 
parts,  and  from  points  on  A  C  draw  lines  converging  to  the  vertex  V.  From  points  on  A  B 
draw  lines  converging  to  the  vertex  V.  The  points  of  intersection  of  these  lines  1  and  1,  2  and 
2,  etc.,  will  be  points  in  the  required  hyperbola.  Produce  the  remaining  branches  of  the  curve 
in  the  same  manner. 

4 


50  CONIC    SECTIONS. 

The  compasses  should  be  used  to  ink  a  small  portion  of  the  curve  on  either  side  of  the  vertex, 
even  if  it  is  possible  to  ink  no  more  than  an  eighth  of  an  inch.  The  remainder  of  the  curve 
requires  the  use  of  a  scroll. 

EXAMPLES. 

1.  Draw  an  ellipse  having  major  axis  6  in.  and  minor  axis  4  in.     Use  First  Method.     Draw  a 

tangent  to  the  curve  at  a  point  2jf  in.  from  minor  axis. 

2.  Draw  an  ellipse  having  major  axis  5|-  in.  and  minor  axis  4-|-  in.     Use  Second  Method.     Try 

also  Third  Method. 

3.  Draw  an  ellipse  having  its  minor  axis  4  in.  and  the  distance  between  foci  3  in.      Use  Third 

Method,  and  test  by  Second  Method. 

4.  Draw  an  ellipse  having  major  axis  6  in.  and  minor  axis  4  in.     Use  Fourth  Method. 

5.  Draw  a  segment  of  an  ellipse  having  an  axis  of  4  in.  and  a  chord  of  5£  in.  which  intersects 

the  axis  at  1|  in.  from  its  extremity.     Use  Fourth  Method. 

6.  Draw  an  ellipse  having  conjugate  diameters  of  5  in.  and  3£  in.,  one  being  horizontal  and  the 

other  making  an  angle  of  60°  with  the  horizontal. 

7.  Draw  an  ellipse  having  its  major  axis  6  in.  and  the  distance  between  foci  4|    in.      Use 

Second  Method,  and  test  three  points  by  Fourth  Method. 

8.  Draw  an  ellipse  having  major  axis  6  in.  and  minor  axis  3  in.     Use  Third  Method,  and  test 

by  Second  Method. 

9.  Draw  a  parabola  having  the  focus  f  in.  from  vertex.     Draw  a  tangent  to  the  curve. 

10.  Draw  a  parabola  having  the  axis  B  V,  Fig.  72,  3|  in.  and  base  E  G,  5|  in. 

11.  Draw  an  hyperbola  having  axis  V  V,  Fig.  73,  U  in.  and  distance  between  foci,  2J  in. 

12.  Draw  an  hyperbola  having  axis  V  V,  Fig.  74,  1|  in.,  B  V,  If  in.,  and  A  D,  4|  in. 


PROJECTION.  51 

IV. 

PROJECTION. 

ORTHOGRAPHIC  PROJECTION,  or  PROJECTION,  as  it  is  commonly  termed,  is  the  art  of  deline- 
ating an  object  on  two  or  more  planes,  suitably  chosen,  and  generally  at  right  angles  to  each 
other,  so  as  to  exactly  represent  the  form  and  dimensions  of  its  lines  and  surfaces,  and  their 
relations  to  each  other. 

Suppose  it  is  required  to  make  the  projection,  or  mechanical  drawing,  of  a  pyramid  having  a 
rectangular  base,  such 'as  is  shown  in  perspective  by  Fig.  1,  Plate  11.  Conceive  the  object  as 
surrounded  by  transparent  planes,  called  PLANES  OF  PROJECTION,  and  shown  in  perspective  in 
Fig.  2,  by  A  D  E  C,  A  B  G  D,  and  A  B  H  C.  We  may  now  have  three  representations  of  the 
object ;  one  by  looking  through  the  front,  one  through  the  top,  and  one  through  the  side  plane  of 
projection.  These  representations  would  be  correct  projections  of  the  object  if  we  imagine  the 
eye  as  being  directly  opposite  all  of  its  points  at  the  same  time,  so  that  the  rays  of  light  from  the 
points  to  the  eye  be  perpendicular  to  the  plane  of  projection ;  or  we  may  conceive  the  eye  as  at 
an  infinite  distance  from  the  plane.  Therefore  ;  if  from  each  point  of  an  object,  perpendiculars 
be,  drawn  to  the  planes  of  projection,  the  intersection  of  these  perpendiculars  with  the  planes  will  be 
the  required  projections  of  the  points,  and  the  lines  joining  these  points  will  be  the  projections  of 
the  lines  and  surfaces  of  the  object. 

Thus  point  1,  Fig.  2,  is  projected  on  the  front  plane  A  D  E  C  at  the  point  1F,  on  the  top 
plane  A  B  G  D  at  1T,  and  on  the  side  plane  A  B  H  C  at  1s.  In  like  manner  the  points  2,  3, 4, 


52  PROJECTION. 

and  5  are  projected  on  the  three  planes.  The  small  letters  F,T,  and  s,  above  and  to  the  right 
of  the  numbers,  indicate  the  plane  upon  which  the  points  lie,  and  when  these  letters  are  not 
affixed  it  signifies  that  the  point  or  line  itself  is  meant,  and  not  its  projection. 

Since  points  1  and  3  are  projected  on  the  front  plane  by  the  same  perpendicular,  it  follows 
that  they  will  have  a  common  point  for  their  projection,  this  being  designated  as  the  projection 
of  two  points  by  the  figures  1F  3P,  the  first  figure  indicating  the  point  nearer  the  plane.  Observe 
similar  cases  on  the  side  plane. 

In  order  to  represent  these  planes  of  projection  upon  a  plane  surface,  as  a  sheet  of  drawing 
paper,  it  becomes  necessary  for  us  to  revolve  two  of  them  into  the  plane  of  the  other  one,  as 
shown  in  Fig.  3,  which  is  a  reproduction  of  Fig.  2  with  the  perspective  of  the  pyramid  omitted 
and  the  planes  revolved.  Thus  the  top  plane  A  B  G  D  is  revolved  about  the  line  A  D  into  the 
position  A  B"  G'  D,  and  the  side  plane  is  revolved  about  the  line  A  C  into  the  position  A  B'  IF  C. 
By  this  means  we  have  obtained,  upon  a  plane  surface,  three  representations  of  the  object  as 
they  would  appear  on  planes  at  right-angles  to  each  other.  A  good  conception  of  the  relation 
of  the  planes  may  be  obtained  by  cutting  a  piece  of  paper  in  the  form  shown  in  Fig.  3  by 
G'  B"  A  B'  H'  E  G',  and  then  folding  it  on  the  lines  A  D  and  A  C.  This  will  represent  three  sides 
of  a  rectangular  prism,  as  shown  in  the  perspective  view,  which  we  may  regard  as  the  trans- 
parent planes  behind  which  the  object  is  supposed  to  be,  and  upon  which  it  is  to  be  projected. 
It  should  be  observed  that  the  top  view  is  always  above  the  front  view,  and  the  side  view  to  the 
right  or  left  of  the  front  view,  according  as  it  may  be  a  view  of  the  right  or  left  of  the  object. 

The  line  A  D  is  known  as  the  FRONT  Axis  OF  PROJECTION.  The  line  A  B  (shown  in  the 
revolved  positions  by  A  B"  and  A  B'),  is  designated  the  SIDE  Axis  OP  PROJECTION,  and  the  line 
A  C,  the  VERTICAL  Axis  OF  PROJECTION.  For  brevity,  these  lines  are  also  designated  as  the 
Front,  Side,  and  Vertical  axes. 


PROJECTION.  53 

That  representation  which  appears  on  the  top  plane  of  projection  is  called  the  TOP  VIEW,  that 
on  the  front  plane,  the  FRONT  VIEW,  and  that  on  the  side  plane,  the  SIDE  VIEW.  The  view  takes 
its  name  from  the  plane  upon  which  the  representation  is  made,  and  not  from  the  face  of  the 
objects  represented.  For  brevity  these  planes  may  be  designated  by  the  letters  T,  F,  and  S. 

From  the  foregoing,  the  following  laws  are  established :  The  front  and  top  views  of  any  point 
lie  in  the  same  vertical  line.  The  front  and  side  views  of  any  point  lie  in  the  same  horizontal 
line.  The  top  and  side  views  of  any  point  lie  equally  distant  from  the  front  and  vertical  axes  of 
projection. 

Suppose  it  is  required  to  obtain  three  views  of  a  rectangular  pyramid,  as  shown  in  Fig.  4,  having 
given  its  dimensions.  First  draw  the  axes  of  projection  D  B'  and  B"  C.  The  view  included 
within  the  angle  DAG  will  be  the  front  view,  that  within  the  angle  B"  A  D,  the  top  view,  and  that 
within  the  angle  B'  A  C,  the  side  view.  It  is  not  necessary  to  limit  these  planes  by  drawing 
boundary  lines,  as  in  Fig.  3,  since  the  planes  are  supposed  to  be  indefinite  in  extent.  In  general, 
draw  that  view,  first,  about  which  most  is  known.  The  following  order  will  be  pursued  in 
this  problem :  Front,  Side,  and  Top  view.  At  some  convenient  distance  below  A  D  and  to 
the  left  of  A  C,  draw  the  line  1F  2F,'fff  the  length  required  to  represent  the  base  of  the  pyramid. 
At  its  middle  point  erect  a  perpendicular  equal  to  the  required  height,  and  connect  points  1F,  5% 
and  2F.  This  will  complete  the  front  view.  Since  "The  front  and  side  views  of  any  point 
lie  in  the  same  horizontal,"  it  follows  that  the  side  view  of  the  vertex,  point  5,  must  lie  in  the 
horizontal  projecting  line  drawn  through  the  point  5%  and  at  some  convenient  distance  to  the 
right  of  A  C.  The  points  of  the  base  will  be  similarly  projected,  and  since  the  pyramid  is  sym- 
metrical, the  points  2s,  1s,  and  4s,  3s,  will  be  equally  distant  from  the  vertical  centre  line  drawn 
through  5s,  the  line  2s  4s  being  made  of  the  length  required  to  represent  the  depth  of  the  pyramid. 
To  project  points  from  front  and  side  views  to  the  top  view,  it  is  necessary  to  observe,  first,  that 


54  PKOJECTION. 

"The  front  and  top  views  of  any  point  lie  in  the  same  vertical  line;"  second,  that  "The  top 
and  side  views  of  an}7  point  lie  equally  distant  from  the  front  and  vertical  axes  of  projec- 
tion." Therefore,  to  project  any  point,  as  5,  into  the  top  view,  a  perpendicular  to  A  D 
through  5F  must  be  drawn  and  the  point  will  lie  in  this  line,  but  its  position  may  not  be  chosen 
as  before,  since  two  projections  of  a  point  being  given,  the  third  is  fixed,  and  the  distance  of  this 
point  from  the  front  plane  of  projection's  determined  by  the  side  view.  Draw  a  perpen- 
dicular through  the  point  5s  until  it  intersects  the  side  axis  A  B'  at  K,  and  remembering  that  the 
lines  A  B'  and  A  B"  are  one,  revolve  the  point  K  by  describing  the  arc  K  K'  from  the  centre  at 
A.  From  K'  draw  the  horizontal  projecting  line  K'  5\which  will  determine  the  top  view  of  the 
point  by  its  intersection  with  5F  5T.  In  like  manner  the  points  of  the  base  may  be  found,  and  since 
the  vertex  is  connected  with  the  four  corners  of  the  base,  we  shall  have  the  lines  5T  1T,  5T  2T,  5T  3T, 
and  5T  4T  to  represent  these  edges. 

Carefully  observe  the  following :  The  projection  of  a  point  is  always  a.  point.  The  projection 
of  a  line  is  either  a  point  or  a  line.  The  projection  of  a  surface  is  either  a  line  or  a  surface.  To 
illustrate:  The  point  1,  of  the  pyramid,  is  projected  as  a  point  on  each  of  the  planes,  as  shown 
at  lp,  1T,  1s.  The  projection  of  the  line  1  2  is  a  line  on  the  front  and  top  planes,  and  a  point  on 
the  side  plane.  The  surface  1  5  2  is  projected  on  the  front  and  top  planes  at  1F  5F  2F  and  1T  5T  2T, 
but  on  the  side  plane  by  the  line  1s  5s,  or  2s  5s.  In  the  same  manner  the  front  axis  of  projection 
A  D,  may  be  considered  as  the  front  view  of  the  top  plane  of  projection,  or  the  top  view  of  the 
front  plane,  and  the  vertical  axis  A  C,  as  the  front  view  of  the  side  plane,  or  the  side  view  of  the 
front  plane  of  projection. 


GENERAL    INSTRUCTION    FOR   DRAWING;-  WTVT 


GENERAL    INSTRUCTION    FOR   DRAWING. 


The  dimension  of  the  paper  will  be  the  same  as  for  the  GEOMETRICAL  PROBLEMS,  the  margin 
line  enclosing  a  space  10x14  in.  Where  several  problems  are  designed  for  the  same  sheet,  it  is 
desirable  to  separate  them  by  a  fine  inked  line.  Four  sheets  will,  in  general,  be  required  for  each 
plate  of  problems.  All  construction  lines  and  lines  of  the  object  should  be  drawn  very  fine  in 
pencil,  and  no  line  that  has  been  useful  in  the  construction  of  the  drawing  should  be  erased.  In- 
visible lines  of  the  object  are  better  dotted  in  pencil,  that  no  mistake  be  made  in  inking  the  same. 
Draw  no  dimension  lines.  Absolute  accuracy  must  be  used  in  the  construction  of  all  figures. 
Only  lines  of  the  object  are  to  be  inked,  the  visible  lines  being  in  full  and  the  invisible  in  dotted 
lines.  Shade  lines  are  to  be  used  only  in  those  problems  indicated,  and  never  shown  in  pencil. 

There  is  but  one  reason  that  can  justify  the  use  of  the  shade  line,  and  that  is,  added  clearness 
to  the  representation  by  indicating  the  relation  of  the  surfaces  to  one  another.  It  often  adds 
much  to  the  beauty  of  a  drawing,  and  for  that  reason  may  sometimes  be  employed.  By  some 
draughtsmen  the  shade  line  is  never  used,  while  by  others  it  is  always  used,  and  both  are  in  error 
since  no  law  can  be  established  concerning  it,  there  being  times  when  it  is  a  mistake  not  to  use 
it,  and  others  where  it  is  equally  wrong  to  use  it.  The  following  method  is  the  only  one  that  can 
consistently  be  used  in  practice  : 

Draw  shade  lines  for  all  right-hand  and  lower  edges  in  all  views.  Shade  the  lower  right-hand 
quadrant  of  outside  circles  and  the  upper  left-hand  quadrant  of  inside  circles.  Never  shade  the 
intersecting  line  between  visible  planes.  The  additional  width  which  is  given  to  a  shade  line 
should  not  encroach  on  the  surface  which  it  bounds. 

Use  red  ink  for  centre  lines,  making  the  lines  full. 

The  following  order  should  be  pursued  in  the  inking  of  all  drawings  :    Small  circles  and 


56  PROBLEMS    IN    PROJECTION. 

circular  arcs  first,  shading  them  at  the  same  time.  Next,  the  larger  arcs,  fine  horizontal  lines, 
fine  vertical  lines,  other  fine  lines,  and  then  all  shade  lines,  pursuing  the  same  order  as  in  the 
inking  of  fine  lines.  Lastly,  ink  centre  lines.  Should  section  lines  be  used,  they  may  be  drawn 
before  the  centre  lines,  if  dimension  lines  are  not  employed,  otherwise  they  must  be  drawn  last. 


PROBLEMS. 

PLATE  12. 

PROBLEM  1.  —  Locate  the  axes  according  to  the  dimensions  given.  Draw  front  and  top 
views,  and  from  these  obtain  the  side  view.  Draw  the  lines  necessary  for  projecting  the  points,  in 
pencil  only.  Having  completed  the  projection  of  the  object,  it  is  an  excellent  practice  to  number 
in  pencil  the  extremity  of  each  line,  as  in  Fig.  4,  PI.  11,  and  thus  acquire  familiarity  with  Ilic 
different  views  of  each  surface,  line,  and  point.  It  will  also  insure  the  representation  of  each  line 
in  every  view. 

PROBLEM  2. — In  this  and  the  following  problems,  the  student  must  use  his  own  judgment 
in  the  location  of  the  axes  of  projection.  The  front  and  side  views  being  given,  the  top  view  will 
be  drawn  last.  See  that  all  the  lines  of  the  object  are  shown  on  this  view. , 

PROBLEM  3.  —  Note  carefully  the  difference  between  the  top  view  of  this  figure  and  that  of  the 
preceding. 

PROBLEM  4.  —  Draw  the  top  view  first,  then  front  and  side  views.  Remember  to  represent 
the  invisible  lines,  since  it  is  required  in  these  problems  to  represent  three  views  of  every  line  of 
the  object. 

PROBLEM  5.  —  Draw  the  top  view  without  the  aid  of  compasses,  observing  that  it  is  an  equi- 
lateral triangle. 


PROBLEMS    IN    PROJECTION.  57 

PROBLEM  6.  —  This  problem  differs  from  the  preceding  in  being  a  pyramid  instead  of  a  prism. 

PROBLEM  7. — Having  given  the  short  diameter  of  a  hexagon,  the  figure  should  be  drawn 
•without  the  aid  of  compasses. 

PROBLEM  8.  —  Although  the  front  view  alone  is  given,  it  is  better  to  draw  the  top  view  first. 

PROBLEM  9.  —  This  prism  having  a  triangular  hole  from  end  to  end,  will  necessitate  the 
representation  of  several  invisible  lines. 

PROBLEM  10.  —  It  is  required  to  represent  the  preceding  object  when  turned  around  on  its 
base.  This,  while  changing  the  angle  at  which  the  top  view  is  drawn,  does  not  alter  the  relation 
of  the  lines  to  each  other,  and,  therefore,  the  top  view  may  be  copied  from  Problem  9.  It  must 
also  be  observed  that  all  points  of  the  object  retain  their  former  height.  Draw  top  view  first,  and 
then  front  and  side  views.  Use  care  to  represent  the  invisible  lines  of  the  object. 

PROBLEM  11.  — This  object  is"  similar  to  the  preceding,  save  that  the  ends  are  bevelled,  and 
the  triangular  space  does  not  pass  entirely  through  the  prism.  It  must  now  be  noticed  that  the 
drawing  of  the  axes  of  projection  is  not  a  necessity,  and  might  have  been  dispensed  with  earlier 
-were  it  not  useful  in  separating  the  planes  of  projection,  and  keeping  clearly  before  the  student 
the  relation  between  the  object  and  the  planes.  The  top  view  of  the  pyramid  in  Fig.  4,  Plate  11, 
could  have  been  drawn  from  the  front  and  side  views,  without  the  use  of  axes,  as  follows  : 
Since  the  axes  are  supposed  not  to  be  drawn,  we  may  draw  the  centre  line  5T  K'  at  any  conveni- 
ent distance  from  the  front  view,  and  from  this  the  lines  1 T  2  T  and  3  T  4  T  may  be  laid  off  on  either 
side,  making  the  distance  between  them  equal  to  2s  4s  on  the  side  view.  This  is  equivalent  to 
considering  the  centre  lines  of  top  and  side  views  as  the  axes.  The  drawing  of  the  projection 
lines  may  also  be  omitted,  as  in  practice  they  would  cause  too  much  confusion  on  the  drawing. 
In  Problem  11,  both  the  axes  and  projection  lines  arc  required  to  be  omitted. 

PROBLEM  12.  —  This  problem  is  similar  to  Problem  10,  save  that  the  projection  lines,  but  not 
the  axes,  are  to  be  omitted. 


58  OBJECTS    OBLIQUE    TO    PLANES. 

OBJECTS    OBLIQUE    TO    PLANES. 

PLATE  13. 

To  enable  the  drawing  of  an  object  to  be  made  when  the  object  is  oblique  to  one  or  more  of 
the  planes  of  projection,  it  is  first  necessary  to  observe  the  changes  which  take  place  in  its  appear- 
ance when  it  is  revolved  in  each  of  three  ways  :  First,  by  turning  it  around  on  its  base.  Second, 
by  revolving  it  forward  or  backward.  Third,  by  revolving  it  to  right  or  left.  The  first  is  a 
revolution  about  a  vertical  axis  of  projection,  or  any  axis  parallel  to  it.  The  second  is  a  revolu- 
tion about  a,  front  axis  of  projection,  or  any  axis  parallel  to  it.  The  third  is  a  revolution  about  a 
side  axis  of  projection,  or  any  axis  parallel  to  it. 

From  Problems  10  and  12  we  have  seen,  that  in  revolving  an  object  about  a  vertical  axis,  the 
top  view  remains  unchanged,  likewise  the  height  of  all  points  of  the  object  is  unaltered,  and  this 
knowledge  was  enough  to  complete  the  necessary  views  of  the  object.  Similarly  by  revolving  it 
about  a  front  axis,  the  side  view  would  be  unchanged,  and  all  dimensions  parallel  to  the  front 
axis,  that  is,  breadths,  would  remain  the  same.  Again,  by  revolving  it  about  a  side  axis,  the  front 
view  would  be  unchanged,  and  all  dimensions  parallel  to  the  side  axis,  viz.,  the  depths,  likewise 
unchanged.  By  the  terms  height,  breadth,  and  depth,  is  not  necessarily  meant  the  height, 
breadth,  and  depth  of  the  object,  but  the  distances  of  points  of  the  object  from  the  planes  of 
projection,  that  is,  distances  parallel  to  the  axes  about  which  they  are  revolved.  To  illustrate, 
suppose  the  object  as  represented  in  Fig.  1,  Plate  13,  be  revolved  about  a  side  axis,  as  shown  in 
Fig.  2.  The  front  view  will  be  unchanged  and  may  be  copied  from  Fig.  1.  To  obtain  the  top 
view,  it  is  only  necessary  for  us  to  remember  that  during  the  revolution,  the  distance  of  each 
of  the  points  from  the  front  plane  of  projection  remains  the  same ;  therefore,  by  projecting  the 
points  from  the  front  view  to  the  top,  and  making  the  distances  from  the  front  axis  the  same  as 


PROBLEMS    IN    PROJECTION.  59 

in  Pig.  1,  all  points  will  be  determined.  From  these  two,  the  side  view  may  readily  be  drawn. 
Fig.  3  represents  the  pyramid  revolved  about  a  side  and  vertical  axis,  but  since  it  is  necessary  to 
perform  the  problems  separately,  Fig.  2  must  first  be  obtained  and  the  pyramid  revolved  from 
this  position  about  a  vertical  axis,  as  was  done  in  Problems  10  and  12.  Fig.  4  represents  the 
pyramid  revolved  backward  about  a  front  axis.  The  term  backward  signifies  a  motion  from  the 
front  plane.  Fig.  5  is  a  revolution  about  a  side  and  front  axis.  Fig.  6  is  a  revolution  about  a 
side,  front,  and  vertical  axis,  the  operations  being  shown  by  Figs.  1,  2,  5,  and  6.  Thus  it  is  to  be 
observed  that  there  is  always  one  view  unchanged,  and  one  set  of  dimensions  unchanged.  The 
following  statement  comprises  all  that  has  been  said  concerning  the  revolution  of  an  object. 

The  unchanged  view  lies  on  that  plane  of  projection  which  is  perpendicular  to  the  axis  of  revo- 
lution, and  the  unchanged  dimensions  are  parallel  to  the  axis  of  revolution. 

In  performing  the  problems,  first  determine  and  draw  the  unchanged  view.  Having  a  set  of 
dimensions  parallel  to  one  of  the  axes,  it  is  then  possible  to  obtain  a  second  view,  and  from  these 
two,  a  third.  Do  not  copy  this  plate,  but  observe  the  principles  in  performing  the  following 
problems. 

PROBLEMS. 

PLATE  12. 

PROBLEM  13.  — Fig.  1  represents  a  hollow  triangular  prism  which  is  required  to  be  shown  in 
three  positions.  In  Fig.  2  it  is  to  be  revolved  about  a  side  axis.  The  front  view,  being  un- 
changed, is  copied  from  Fig.  1,  as  shown.  The  points  may  now  be  projected  into  the  top  view, 
and  since  the  depths  remain  the  same,  the  points  will  have  the  same  relative  position  with  respect 
to  the  front  plane  of  projection.  In  Fig.  3  the  prism  is  revolved  about  a  front  axis,  the  side  view 
being  unchanged.  Fig.  4  is  a  revolution  about  a  vertical  axis. 

The  different  figures  may  be  separated  by  pencil  lines  only,  and  the  shade  lines  omitted  in 
these  problems. 


60  PROBLEMS    IN    PROJECTION. 


PLATE  14. 

PROBLEM  14.  —  This  problem  is  similar  to  the  preceding,  but  in  this  case  the  unchanged  view 
is  not  shown  on  the  plate.  Use  care  to  revolve  the  object  in  the  direction  prescribed,  and  through 
the  proper  number  of  degrees. 

PROBLEM  15.  —  Having  drawn  the  pyramid,  as  shown  in  Fig.  1,  proceed  with  the  problems  in 
the  order  indicated  by  the  figures.  Just  here  it  is  well  to  note  that  lines  of  the  object  which  are 
parallel,  are  always  parallel  in  projection.  In  many  cases  it  will  be  found  helpful  to  number  the 
points,  as  in  Plate  3,  but  should  this  be  done,  great  care  must  be  used  to  retain  the  same  number 
for  each  point  throughout  the  problem. 

PROBLEM  16.  —  This  problem  differs  from  the  preceding  riot  only  in  that  the  object  to  be 
revolved  is  a  cone,  but  because  of  the  absence  of  a  view  from  which  the  necessary  dimensions  for 
the  top  view  may  be  obtained.  In  this  case  the  front  view  may  be  as  readily  drawn  as  though 
the  cone  was  resting  on  its  base,  but  in  the  top  view,  the  depths  are  apparently  missing.  In 
order  to  find  the  curve  of  the  base,  in  the  top  view  and  the  several  revolved  positions,  we  must 
consider  it  as  consisting  of  points,  whose  positions  are  to  be  determined  in  the  same  manner 
as  the  corners  of  the  base  of  a  pyramid.  The  points  E  F,  A  F,  C  F,  F F,  D F,  BF,  are  points  of  the  base 
whose  position  in  the  top  view  being  found,  the  required  curve  can  be  drawn  through  them. 
Remembering  the  base  of  the  cone  to  be  a  circle,  imagine  that  half  of  it  ,lying  next  the  front 
plane  of  projection,  to  be  revolved  about  the  diameter  of  the  base  parallel  to  that  plane,  into  the 
position  EFA'C'FF.  The  point  which  was  at  A  F  is  now  revolved  to  A',  and  the  point  CFto  C'. 
Next  conceive  a  vertical  plane  as  passing  through  the  axis  of  the  cone  and  this  same  diameter  of 
the  base.  The  top  view  of  this  plane  would  be  the  line  E  T  G  T,  and  the  distance  of  the  point 
AT  from  this  plane  will  equal  AFA'.  The  distance  of  the  point  BT  on  the  other  side  would  also 


REVOLUTION    OF    LINES.  61 

equal  A  F  A'.  The  points  C  T  and  D  T  would  be  at  a  distance  from  this  plane  equal  to  C  FC',  and 
the  points  ET  and  FT  lie  in  the  plane.  In  like  manner  obtain  a  sufficient  number  of  points  to 
enable  the  curve  to  be  drawn.  Since  the  vertex  lies  in  the  plane  ETGT,  its  projection  may  be 
obtained  at  G  T,  and  by  drawing  tangent  lines  from  this  point  to  the  curve  of  the  base  the  figure 
is  completed.  Next  revolve  the  cone  as  directed,  determining  the  points  in  the  same  manner  as 
though  they  were  the  corners  of  a  polygon. 

All  the  curves  in  this  problem,  being  ellipses,  might  have  been  obtained  by  first  finding  the 
major  and  minor  axes  and  constructing  the  curves  from  them.  While  this  would  be  a  shorter 
method,  it  would  fail  to  give  the  practice  most  necessary  at  this  stage,  beside  involving  problems 
not  yet  explained. 

PROBLEM  17.  —  The  revolution  of  a  surface  about  the  several  axes  of  projection.  Solve  as  in 
Problem  15. 

SPECIAL,   METHODS    FOR   THE    REVOLUTION    OF   LINES,    SURFACES,    AND    SOLIDS. 

PLATE  15. 

The  methods  used  in  the  foregoing  problems  have  been  such  as  would  best  illustrate  and 
explain  the  fundamental  principles  of  projection,  and  their  application  has  been  chiefly' applied  to 
the  representation  of  solids,  as  being  more  easily  comprehended  than  that  of  lines  and  surfaces. 
Some  of  the  shorter  methods  for  the  solution  of  those  problems,  together  with  the  consideration 
of  the  projection  of  lines,  their  measurement,  and  the  determining  of  the  angles  which  they  make 
with  the  planes  of  projection,  are  treated  in  the  following : 

FIRST  METHOD.  —  Let  A F  B F,  A  T  B T,  and  A s  Bs,  Fig.  1,  Plate  15,  represent  three  views  of  a 
line  inclined  to  each  of  the  planes  of  projections,  which  latter  will  for  brevity  be  designated  by 
F,  T,  and  S.  It  is  required  to  find  the  true  length  of  this  line,  and  the  angles  which  it  makes 


62  REVOLUTION    OF    LINES. 

with  the  planes  of  projection.  Since  the  line  is  inclined  to  each  of  these  planes,  its  projection  on 
them  will  be  foreshortened  views  of  the  line.  In  order  that  the  line  shall  be  parallel  to  one  of  the 
planes,  and  thus  seen  in  its  true  length  on  that  plane,  it  will  be  necessary  to  revolve  it  about  one 
of  the  axes  according  to  the  principles  already  established  for  the  revolution  of  an  object.  It 
may  be  revolved  about  a  vertical  axis  until  it  becomes  parallel  to  either  F  or  S  ;  or  about  a  front 
axis  until  parallel  to  either  F  or  T  ;  or  about  a  side  axis  until  parallel  to  S  or  T.  Fig.  2  repre- 
sents the  line  revolved  about  a  vertical  axis  and  parallel  to  F,  the  revolved  position  being  shown 
by  the  broken  line  AF  B'.  The  top  view  of  the  line,  which  in  the  revolution  about  a  vertical  axis 
we  know  to  be  unchanged,  is  first  drawn  parallel  with  the  front  axis  of  projection,  in  which  posi- 
tion it  will  be  parallel  to  F.  The  height  being  unchanged,  the  front  view  may  next  be  obtained, 
and  the  representation  will  be  that  of  a  line  parallel  to  F,  and  therefore  seen  in  its  true  length  on 
that  plane.  This  view  also  shows  the  correct  angle  R,  which  the  line  makes  with  a  horizontal 
plane,  since  its  position  relative  to  that  plane  is  unchanged  by  the  revolution. 

Similarly  the  line  may  be  revolved  about  a  side  axis  until  it  is  parallel  with  T,  as  in  Fig.  3, 
when  its  true  length  may  be  measured  on  the  top  view.  This  view  also  gives  the  angle  which 
the  line  makes  with  F,  or  any  plane  parallel  thereto. 

Fig.  4  illustrates  the  revolution  of  the  line  about  a  vertical  axis  until  parallel  to  S.  This  is 
seldom  used,  as  it  is  much  more  simple  to  obtain  the  true  length  on  F,  as  in  Fig.  2,  and  avoid 
the  drawing  of  an  extra  view. 

SECOND  METHOD.  —If  the  quadrilateral  formed  by  a  line,  the  projecting  lines  of  its  extremi- 
ties, and  its  projection,  be  revolved  about  the  latter  until  the  surface  coincides  with  the  plane  of 
projection,  the  revolved  position  of  the  line  itself  will  exactly  represent  the  length  of  the  line,  and 
the  angle  which  it  makes  with  the  plane  of  projection  will  be  shown  on  the  plane  into  which  it 
has  been  revolved. 


REVOLUTION    OF    LINES.  63 

Iii  Fig.  5  let  AT  BT  be  the  horizontal  projection  of  a  line,  and  A  F  C,  BFD,  the  vertical  pro- 
jection of  the  projecting  lines  of  its  extremities.  Since  these  last  named  lines  are  seen  in  their  true 
length  in  this  view,  we  have  three  sides  of  the  quadrilateral,  and  the  fourth  will  be  the  true  length 
of  the  line.  To  obtain  this  result,  draw  ATC'  and  B  T  D'  equal  to  A  F  C  and  B  D,  respectively, 
and  perpendicular  to  A  T  B T.  The  points  C',  D',  will  be  the  extremities  of  the  revolved  position  of 
the  required  line,  and  the  angle  which  C'  D'  makes  with  AT  BT  is  the  angle  which  the  given  line 
makes  with  T,  or  any  horizontal  plane.  The  plane  in  which  this  quadrilateral  lies  is  known  as 
the  horizontal  projecting  plane  of  the  line,  and  AT  BT  is  called  its  horizontal  trace,  it  being  the 
line  in  which  the  projecting  plane  intersects  the  horizontal  plane. 

Similarly,  in  Fig.  6,  the  vertical  projecting  plane  of  the  line  has  been  used,  and  the  revolution 
made  into  F,  about  its  vertical  trace,  A  F  B  F,  which  is  the  vertical  projection  of  the  given  line. 
The  true  length  of  the  line  is  C'  D',  and  the  angle  which  this  line  makes  with  A  p  BF  is  the  angle 
which  the  given  line  makes  with  F. 

An  application  of  the  foregoing  principles  is  illustrated  by  Fig.  7.  This  represents  a  rectan- 
gular surface  with  its  long  edges  making  angles  of  0°,  30°,  60°,  and  90°  with  a  vertical  plane,  and 
35°  with  a  horizontal  plane.  This  surface  might  first  have  been  drawn  with  its  long  edges  hori- 
zontal and  making  the  required  angles  with  the  vertical  plane,  after  which  it  could  have  been 
revolved  about  a  side  axis  until  the  edges  made  the  required  angle  of  35°  with  the  horizontal 
plane.  This  would  have  necessitated  the  drawing  of  two  more  views,  which  may  be  avoided  in 
the  following  manner.  Suppose  the  surface  to  coincide  with  F,  as  shown  by  AFBFDFC%  the 
long  edges  making  the  required  angle  with  T.  Next,  conceive  the  surface  as  being  revolved 
about  one  of  its  short  edges,  as  AF  C  F,  until  the  required  angles  with  F  are  obtained.  Suppose 
one  such  angle  to  be  30°,  draw  A  F  B'  equal  to  A  F  BF,  and  making  an  angle  of  30°  with  it.  This 
will  represent  the  position  of  the  required  line  when  revolved  about  the  trace  of  its  vertical  pro- 


64  KEVOLUTION    OF    LINES. 

jecting  plane,  precisely  as  was  done  in  Fig.  6.  Next,  revolve  the  line  about  this  trace  and  into 
the  required  position,  obtaining  thereby  the  points  BnF  and  DnF.  The  horizontal  projection  of  these 
points  may  now  be  obtained,  since  their  distance  from  F  is  determined,  being  equal  to  B',  Bi.F.  In 
like  manner  the  other  required  positions  of  the  surface  may  be  obtained. 

Fig.  8  illustrates  the  case  of  a  horizontal  hexagonal  prism,  the  lateral  edges  of  which  make  an 
angle  of  30°  with  F.  In  this  problem  the  top  view  may  readily  be  drawn,  and  from  that  the  pro- 
jection of  the  points  in  the  front  view  determined,  except  as  to  height.  If  one  of  the  bases  of  the 
prism  be  revolved  into,  or  parallel  to,  T,  the  required  measurement  may  be  readily  obtained. 
This  is  similar  to  Problem  16,  Plate  14,  and  like  it,  the  base  might  have  been  revolved  about  its 
horizontal  diameter. 

From  the  foregoing,  the  following  principles  may  be  established :  — 

The  true  length  of  a  line  can  be  seen  only  on  that  plane,  or  planes,  to  ivhich  the  line  is  parallel. 

If  a  right  line  is  perpendicular  to  a  plane  of  projection,  its  projection  on  that  plane  will  be  a 
point. 

If  a  right  line  is  parallel  to  a  plane  of  projection,  its  projections  on  the  other  two  planes  will  be 
either  a  point,  or  a  line  parallel  to  an  axis  of  projection. 

If  a  surface  is  parallel  to  a  plane  of  projection,  its  projection  on  the  other  planes  will  be  a  line 
parallel  to  an  axis  of  projection. 


SPECIAL    PROBLEMS    IN    PROJECTION.  65 

SPECIAL    PROBLEMS    IN    PROJECTION. 

Each  problem  will  require  a  space  of  7x5  in. 

Three  views  are  required,  and  all  invisible  as  well  as  visible  lines  should  be  shown  on 
each  view. 

Leave  all  construction  lines  in  pencil. 

All  polygons  referred  to  are  regular  polygons. 

1.  Draw  the  frustum  of  an  octagonal  pyramid  having  its  base  horizontal  and  two  of  its  edges 
making  an  angle  of  30°  with  F.  The  diameter  of  the  circumscribing  circle  of  its  lower 
base  is  If  in.,  and  of  the  upper  base,  If  in.  The  altitude  is  1|  in. 

'2.     Revolve  the  pyramid  as  required  in  1,  30°  to  right,  about  side  axis. 

3.  Draw  a  pentagonal  prism  resting  on  one  of  its  faces  and  having  its  lateral  edges  at  an 

angle  of  22£°  with  F.  Diameter  of  circumscribing  circle  of  base,  1|  in.  Length  of 
prism,  2|  in. 

4.  Draw  an  equilateral  triangular  prism  resting  on  one  of  its  faces,  and  its  lateral  edges  making 

an  angle  of  15°  with  F.  The  edges  of  the  base  are  1|  in.,  and  the  length  of  the  prism, 
2-£  in.  There  is  an  equilateral  triangular  hole  extending  through  the  bases  and  making 
the  thickness  of  the  sides  |  in. 

5.  Draw  a  cylinder  with  its  axis  parallel  to  F,  and  at  an  angle  of  60°  with  T.     The  diameter  of 

the  base  is  If  in.,  and  the  length  of  cylinder,  2^  in.  Obtain  the  ellipses  by  the  method  of 
trammels. 

6.  Draw  an  equilateral  triangular  pyramid  having  an  altitude  of  2J  in.,  and  the  edges  of  the 

base,  If  in.  The  base  makes  an  angle  of  30°  with  T,  and  one  of  its  edges  is  perpen- 
dicular to  F.  5 


66  SPECIAL    PROBLEMS    IN    PROJECTION. 

7.  Revolve  the  pyramid  as  required  in  6,  45°  forward. 

8.  Draw  a  box  having  the  following  outside  dimensions.     Length,  2  in.,  width,  If  in.,  depth, 

including  cover,  1  in.     Thickness  of  material,  \  in.     The  long  edges  of  the  box  are  hori- 
zontal, and  make  an  angle  of  30°  with  F.   The  cover  is  hinged  on  long  edge,  and  opened  30°. 

9.  Draw  a  pyramid  formed  of  four  equilateral  triangles  having  2|  in.  sides.     The  base  is 

horizontal,  and  one  of  its  edges  makes  an  angle  of  30°  with  F. 

10.  Draw  a  regular  octahedron  having  edges  If  in.  long.     The  square  surface  to  be  horizontal, 

with  two  of  its  edges  making  angles  of  30°  with  F. 

11.  Draw  a  rectangular  surface,  I£x2f  in.,  in  the  following  positions.     The  short  edges  hori- 

zontal, and  making  an  angle  of  75°  with  F  ;   the  long  edges  making  angles  of  15°,  30°, 
and  45°  with  T. 

12.  Revolve  the  surface  from  the  positions  required  in  11,  15°  forward. 

13.  Draw  an  isosceles  triangle  in  three  positions  as  follows.     The  base  lying  on  F,  and  inclined 

at  an  angle  of  30°  with  T.     The  altitude  making  angles  of  90°,  30°,  and  15°  with  F.     The 
base  of  triangle  is  If  in.,  and  the  altitude,  2^  in. 

14.  Draw  the  same  triangle  revolved  from  the  positions  in  13,  30°  in  either  direction  about  a 

vertical  axis. 

15.  Draw  an  isosceles  triangle  in  the  following  positions.     The  base  horizontal,  and  making  an 

angle  of  60°  with  F.     The  altitude  making  angles  of  45°  and  60°  with  T.     The  base  of 
triangle  is  2  in.,  and  the  altitude,  2-|  in. 

16.  Revolve  the  same  triangle  from  the  positions  in  15,  30°  backward  about  front  axis. 

17.  Draw  an   octagonal  surface  inclined   at  an  angle   of  60°  with  T,  two  of  its  edges  being 

horizontal,  and  making  angles  of  15°  with  F.     The  diameter  of  circumscribing  circle 
is  2.i  in. 


SPECIAL    PROBLEMS    IN    PROJECTION.  67 

18.  Draw  an  hexagonal  surface  inclined  at  an  angle  of  45°  with  F,  two  of  its  edges  being  parallel 

to  F,  and  making  angles  of  30°  with  T.     The  long  diameter  of  hexagon  is  2|-  in. 

19.  Draw  the  projections  of  a  line  located  as  follows.     The  left-hand  extremity  of  the  line  is 

|  in.  behind  F,  and  \  in.  below  T.  The  right-hand  extremity  is  1|  in.  behind  F,  and 
If  in.  below  T.  The  horizontal  projection  of  the  line  makes  an  angle  of  30°  with  F. 
Find  the  length  of  the  given  line  by  revolving  it  parallel  to  each  of  the  planes  of 
projection. 

20.  Draw  the  projections  of  a  line  of  which  the  left-hand  extremity  is  1  in.  behind  F,  and  If  in. 

below  T.  The  right-hand  extremity,  f  in.  behind  F,  and  f  in.  below  T.  The  horizontal 
projection  making  an  angle  of  15°  with  F.  Find  the  length  of  the  line  by  revolving  it 
into  the  planes  of  projection  by  the  second  method,  page  62. 


68  THE    DEVELOPMENT    OF    SURFACES. 

j 

V. 

THE    DEVELOPMENT    OF    SURFACES. 

PLATE  16. 

IT  is  often  required  to  illustrate  the  surfaces  of  an  object  in  such  a  manner  that  a  pattern 
being  made  from  it,  and  properly  folded  or  rolled,  would  exactly  reproduce  the  object.  In 
order  to  do  this,  an  outline  of  each  surface  must  be  obtained,  as  it  would  appear  on  a  plane  of 
projection  parallel  to  it,  that  is,  there  would  be  no  foreshortening  of  the  surface.  Fig.  1  is  the 
projection  of  a  square  pyramid,  and  it  is  required  to  produce  the  pattern,  which  if  properly  folded, 
would  make  a  pyramid  like  the  one  in  the  drawing.  This  operation  is  called  the  development  of 
the  surface.  Since  four  of  the  surfaces  are  triangles,  it  is  possible  to  obtain  their  true  area  by 
finding  the  length  of  their  sides.  A  line  is  seen  in  its  true  length  on  a  plane  when  it  is  parallel  to 
that  plane.  Thus  the  line  D  E  may  be  measured  on  the  front  view,  because  the  line  of  the  pyramid 
which  it  represents  is  parallel  to  that  plane,  and  we  know  that  it  is  parallel  to  the  plane  because 
the  top  view  of  the  line  is  parallel  to  the  front  axis  of  projection.  Neither  of  the  lines  E  A  orE  C 
may  be  measured  from  the  drawing,  but  since  the  base  of  the  pyramid  is  symmetrical  with  respect 
to  the  axis,  we  know  these  lines  or  edges  to  be  of  equal  length  with  D  E  and  E  B.  The  only 
undetermined  line  of  the  surface  A  E  D  is  A  D,  which  may  be  measured  on  the  top  view,  it  being 
a  horizontal  line.  Fig.  2  shows  these  lines  in  their  proper  lengths  and  relation  to  each  other. 
The  surfaces  A  E  B,  B  E  C,  and  C  E  D  are  of  the  same  shape  and  size  as  A  E  D,  and  may  be 
drawn  in  connection  with  it.  The  bottom  surface  alone  remains  to  be  drawn,  and  this  being 


THE    DEVELOPMENT    OF    SURFACES.  69 

parallel  to  the  top  plane,  is  already  seen  in  its  true  size  and  shape,  and  may,  therefore,  be  copied 
directly.  Having  the  length  of  one  of  the  edges  we  might  have  used  it  for  a  radius  to  describe 
the  arc  D  A  B  C  D,  and  by  spacing  off  the  bottom  edges  D  A,  A  B,  etc.,  have  attained  the  same 
result  in  an  easier  and  more  practical  manner. 

Fig.  3  illustrates  a  rectangular  pyramid  which  it  is  required  to  develop,  after  removing  such 
portion  of  the  top  as  is  indicated  by  the  line  FF  KF.  The  operations  are  as  follows :  First  obtain 
the  top  view,  showing  the  top  removed,  and  similarly  the  side  view,  if  required.  Secondly,  obtain 
the  development  of  the  entire  pyramid,  disregarding  the  cutting  plane.  Finally,  determine  that 
portion  of  the  developed  surface  not  removed  by  the  cutting  plane,  to  which  must  be  added  the 
section  cut  by  the  plane.  As  the  first  operation  is  sufficiently  well  indicated  by  the  drawing,  we 
will  consider  only  the  development.  None  of  the  inclined  edges  being  parallel  to  either  of  the 
planes  of  projection,  it  is  necessary  to  revolve  one  of  these  edges  until  it  shall  become  parallel  to 
a  plane,  when  it  will  be  possible  to  measure  it  on  that  plane.  Let  A  E  be  the  line  to  be  revolved. 
Since  this  is  a  revolution  about  a  vertical  axis,  the  top  view  of  the  line  will  be  changed  in  position 
but  not  in  length,  and  will  be  shown  by  A'  ET .  The  front  view  will  then  be  A"  EF  which 
represents  the  true  length  of  the  line.  Since  all  the  inclined  edges  are  of  the  same  length,  with 
radius  equal  to  A"  EF,  describe  an  arc  on  which  the  chords  A  B,  B  C,  C  D,  and  D  A  may  be 
drawn,  as  shown  in  Fig.  4,  their  lengths  being  obtained  from  the  top  view.  The  development 
of  the  base  is  obtained  directly  from  the  top  view. 

The  surface  B  E  C  can  be  obtained  in  another  manner,  as  follows :  Since  the  true  size  of  a 
surface  is  always  to  be  found  on  a  plane  to  which  it  is  parallel,  we  have  only  to  draw  a  plane, 
X  Y,  parallel  to  this  surface,  and  project  on  to  it  in  order  to  obtain  the  required  surface.  This 
new  plane  is  perpendicular  to  the  front  plane,  but  not  to  the  other  planes.  The  line  X  Y  may  be 
regarded  as  the  axis  of  the  new  plane,  which,  being  revolved  to  coincide  with  the  front  plane, 


70  THE    DEVELOPMENT    OF    SURFACES. 

would  be  shown  as  in  the  figure,  the  distances  of  the  points  E',  B',  and  C"  from  the  line  X  Y  being 
their  distances  from  the  front  plane  of  projection.  As  the  centre  line  E'  S  is  known  to  be  parallel 
to  the  front  plane  of  projection,  it  is  unnecessary  to  draw  the  axis  X  Y,  since  measurements  may 
be  made  equally  well  from  this  centre  line. 

Finally,  having  obtained  the  development  of  the  entire  pyramid,  it  is  required  to  find  the 
length  of  the  edges  when  cut  off,  and  the  section  made  by  the  cutting  plane.  Since  we  have 
found  it  possible  to  obtain  the  true  length  of  the  inclined  edges,  we  may  in  like  manner  find  that 
portion  of  them  included  between  the  base  and  cutting  plane.  As  A"  EF  may  be  considered  as 
the  revolved  position  of  any  one  of  these  lines,  and  since  the  heights  remain  unchanged  by 
revolving  about  a  vertical  axis,  the  true  length  of  AF  KF  and  DF  HF  will  be  A"  N,  and  the 
length  of  BF  FF  and  CF  GF  will  be  A"  0.  (It  is  generally  better  to  lay  off  these  distances 
from  the  apex  instead  of  the  base.)  The  cut  surface  F  G  H  K  should  be  found  in  the  same 
manner  as  the  surface  E'  B'  C',  observing  that  the  lines  FT  GT  and  HT  KT  may  be  measured 
on  the  top  view  as  these  lines  are  horizontal.  A  copy  of  this  surface  should  be  drawn  in 
connection  with  the  developed  surface  as  shown  in  Fig.  4. 

PROBLEMS. 

PLATE  17. 

Tn  performing  these  problems,  pursue  the  order  specified  in  the  instructions.  Three  views 
and  the  complete  development  will  be  required  in  each  case.  In  general,  begin  to  develop  the 
surface  on  the  shortest  edge.  It  will  assist  the  student  to  a  better  understanding  of  this  subject  if 
the  developed  surface  be  copied  on  to  stiff  paper  and  afterwards  cut  out  and  folded.  If  this  be 
done,  a  smnll  lap  should  be  left  on  some  of  the  edges,  to  fasten  them  together.  Shade  lines  must 
not  be  used  on  the  development. 


THE  DEVELOPMENT  OF  SURFACES.  71 

PROBLEM  18. —  This  is  a  prism  having  a  square  base.  The  vertical  edges  being  parallel  to 
the  front  plane,  are  seen  in  their  true  length  on  the  front  view.  The  length  of  the  edges  of  the 
base  can  be  obtained  from  the  top  view,  and  the  true  shape  of  the  upper  base  will  be  found  by 
projecting  it  on  to  an  auxiliary  plane,  as  shown  in  Plate  16. 

PROBLEM  19. —  This  differs  from  the  preceding  only  in  being  an  hexagonal  prism. 

PROBLEM  20. —  In  this  and  the  following  problems,  only  those  lines  should  be  inked  which 
lie  below  the  cutting  plane,  the  upper  portion  being  supposed  to  be  removed. 

PROBLEM  21. —  This  is  a  pyramid  having  a  square  base  the  long  diameter  of  which  is  given. 

PROBLEM  22. —  The  cutting  plane  makes  an  angle  of  50°  in  this  case,  and  the  auxiliary  plane 
must  be  at  the  same  angle,  the  projecting  lines  being  drawn  perpendicular  to  it. 

PROBLEM  23. —  The  surface  cut  by  the  plane  will  not  be  symmetrical  with  respect  to  the 
centre  line,  as  in  the  previous  problems.  Care  must,  therefore,  be  used  to  take  no  dimensions 
from  the  top  view  that  may  not  be  represented  by  horizontal  lines. 

PROBLEM  24. —  This  is  similar  to  Problem  23,  but  one  of  the  inclined  edges  will  not  appear 
in  the  finished  development,  as  it  is  entirely  removed  by  the  cutting  plane.  Begin  to  develop  the 
surface  on  this  line. 

PROBLEM  25.—  In  the  preceding  problems,  the  inclined  edges  of  the  pyramids  have  been  of 
the  same  length,  and  having  determined  the  revolved  position  of  one  of  them,  it  sufficed  for  the 
others.  The  form  of  this  object  makes  it  necessary  to  obtain  them  separately.  One  of  these 
edges,  A  D,  is  shown  in  its  revolved  position  at  A  D'.  One-half  of  the  development  is  also  shown, 
and  the  method  and  order  for  drawing  the  lines  indicated  by  the  numbers.  Thus,  the  line  A  B  is 
drawn  first,  and  then  arcs  2  and  3  described  from  its  extremities,  A  and  B,  with  radii  equal  to  the 
true  lengths  of  A  C  and  B  C,  which  determines  the  point  C.  In  like  manner  the  remaining  points 
and  lines  are  found. 


72  THE    DEVELOPMENT    OF    SURFACES. 


DEVELOPMENT    OF    SURFACES    OF   REVOLUTION. 

A  surface  may  be  conceived  to  be  generated  by  the  motion  of  a  line.  Such  a  line  is  called 
the  GENERATRIX,  the  path  described  by  one  of  its  points  is  called  the  DIRECTRIX,  and  the  different 
positions  of  the  GENERATRIX  are  called  ELEMENTS. 

Suppose  a  right  line  to  have  a  motion  about  another  right  line,  known  as  the  Axis,  from  which  it 
is  always  equidistant,  and  to  which  it  is  parallel,  then  will  the  successive  positions  of  this 
generating  line  constitute  the  surface  of  a  cylinder.  If  one  end  of  the  generatrix  were  fixed  to 
the  axis,  and  the  other  free  to  describe  a  circular  path,  the  surface  generated  would  be  that  of  a 
cone.  These  surfaces  are  called  surfaces  of  revolution,  and  may  be  regarded  as  consisting  of  an 
infinite  number  of  lines,  or  elements,  which  in  the  first  case  are  parallel  to,  and  in  the  second  case 
intersect,  the  axis.  This  conception  of  the  cylinder  and  cone  is  necessary  to  the  study  of  the 
development  and  intersection  of  surfaces.  (See  page  26.) 

We  may  also  imagine  the  cylinder  as  generated  by  the  motion  of  a  circle  whose  directrix  is  a 
right  line. 

THE    CYLINDER. 

PLATE  18. 

Three  views  of  a  cylinder  are  represented  by  Fig.  1,  and  from  these  it  is  required  to  develop 
the  cylinder.  Assume  a  number  of  elements,  and,  for  convenience,  they  should  be  equidistant, 
These  may  be  employed  to  obtain  other  views,  sections,  and  development,  in  precisely  the  same 
manner  as  though  they  were  the  edges  of  a  prism,  save  that  instead  of  connecting  their 
extremities  by  right  lines,  a  curve  must  be  drawn  through  them.  It  will  be  observed  that  in  this 
problem  the  revolved  section  is  an  ellipse,  the  major  axis  of  which  is  equal  to  AF  CF,  and  minor 


THE  DEVELOPMENT  OF  SURFACES.  73 

axis  equal  to  the  diameter  of  the  cylinder.  From  this  data  the  curve  might  be  drawn,  but  it  is 
better  to  use  this  method  as  a  test  for  the  ellipse  after  having  obtained  the  curve  by  means  of  the 
elements.  To  develop  the  cylinder  [Fig.  2],  first  obtain  the  circumference  of  the  base  by  com- 
puting the  same  from  the  diameter,  and  having  laid  it  off,  as  would  be  done  in  the  case  of  a  prism, 
divide  it  into  as  many  parts  as  there  are  elements.  The  perpendiculars  to  the  base  drawn 
through  these  points  will  be  the  required  elements,  and  their  lengths  may  be  obtained  directly 
from  the  front  view,  since  they  are  parallel  to  the  front  plane.  A  free-hand  curve  should  be 
carefully  pencilled  through  these  points,  and  afterwards  neatly  inked  by  the  aid  of  compasses 
and  scrolls. 

THE    CONE. 

PLATE  18. 

Fig.  3  illustrates  a  cone  cut  by  vertical  and  oblique  planes.  Draw  the  requisite  number  of 
elements  by  dividing  the  circle  of  the  base  in  the  top  view,  and.  projecting  these  points  into  the 
front  view,  as  at  BF,  EF,  GF,  HF,  etc.  These  are  then  connected  with  the  vertex  AF.  It  is  now 
possible  to  obtain  the  top  view  of  the  section  made  by  the  cutting  plane  X  Y,  in  the  same  manner 
as  though  the  cone  were  a  pyramid  having  these  elements  for  its  edges.  A  more  simple  and 
accurate  method  is  as  follows :  Through  the  front  view  of  any  point  NF,  lying  on  the  surface  of 
the  cone,  and  also  in  the  cutting  plane,  draw  a  horizontal  line  which  will  represent  the  front  view 
of  a  circle  of  the  cone,  and  shown  on  the  top  view  by  the  fine  dotted  circle  drawn  through  NT. 
Since  the  assumed  point  must  lie  on  this  circle,  as  well  as  on  its  vertical  projecting  line,  it  must 
lie  at  their  intersection  NT.  Of  course  it  is  necessary  to  draw  only  small  arcs  to  intersect  the 
projecting  line.  In  like  manner  any  number  of  points  may  be  obtained,  and  through  them,  the 
curve  described.  If  the  elements  of  the  cone  have  already  been  drawn,  as  would  be  necessary  if 


74  THE    DEVELOPMENT    OF    SURFACES. 

it  is  to  be  developed,  the  points  assumed  had  better  be  at  the  intersection  of  the  cutting  plane  and 
the  elements.  This  curve,  being  an  ellipse,  may  be  obtained  by  finding  the  major  and  minor 
axes,  and  on  these  constructing  the  curve ;  but  as  in  the  case  of  the  cylinder,  this  method  should 
be  used  only  as  a  test  of  the  ellipse,  until  the  problem  is  thoroughly  understood.  The  side  view 
illustrates  the  true  section  made  by  the  vertical  cutting  plane,  since  it  is  parallel  to  the  side 
plane  of  projection.  The  top  view  of  this  section  is  a  straight  line,  but  it  is  just  as  necessary  to 
find  the  points  by  projection,  as  in  the  case  of  the  ellipse,  since  their  distances  on  either  side  of 
the  axis  must  be  known  in  order  to  obtain  the  side  view.  Thus  the  top  view  of  point  0  must  lie 
in  the  vertical  projecting  line  drawn  through  0F,  and  also  in  a  circle,  the  diameter  of  which  is 
equal  to  V  M,  therefore  at  0T.  From  this,  the  side  view  of  the  point  may  be  obtained,  its  dis- 
tance, Qs  0s,  from  the  axis  being  equal  to  QT  0T. 

Proceed  with  the  development,  as  in  the  case  of  the  pyramid,  observing  that  because  all  the 
elements  are  equal  in  length,  the  development  of  the  base  will  be  a  circular  arc  of  length  equa.l 
to  the  circumference  of  base,  and  radius  equal  to  the  length  of  an  element.  Having  divided  this 
arc  into  as  many  parts  as  there  are  elements,  proceed  to  draw  them,  after  which  the  true  length 
of  the  cut  portion  may  be  found  as  follows  :  Having  drawn  any  element,  as  A  G,  Fig.  4,  it  is 
required  to  find  the  points  P  and  R.  Since  the  line  A  G,  upon  which  they  lie,  is  not  seen  in  its 
true  length  in  any  of  the  views,  it  must  be  revolved  parallel  to  one  of  the  planes.  AF  BF  will 
represent  its  position  when  revolved  parallel  to  the  front  plane.  The  point  PF  will  be  seen  at 
P',  and  RF  at  R',  and  the  lengths  Ap  P',.  and  P'  R',  may  be  laid  off  on  the  line  A  G  of  the 
developed  surface.  Similarly  other  points  are  found.  In  making  a  model,  as  was  advised  in  the 
case  of  the  prism  and  pyramid,  it  will  of  course-  be  necessary  to  add  the  revolved  sections,  and 
the  base,  in  order  to  complete  the  development. 


THE    DEVELOPMENT    OF    SURFACES.  75 

PROBLEMS. 

PLATE  19. 

PROBLEM  26. —  The  development  of  a  cylinder  is  required.  Leave  all  elements  in  pencil,  and 
use  great  care  in  the  construction  and  inking  of  the  curves. 

PROBLEM  27. —  The  development  of  the  cone  will  require  even  more  care  than  that  of  the 
cylinder.  Use  the  second  and  more  simple  method  for  determining  the  top  view  of  points 
made  by  the  cutting  plane.  Remember  to  revolve  the  elements  before  measuring. 

PROBLEM  28. —  This  problem  is  introduced  as  a  review  of  conic  sections  [page  43],  and  as  a 
test  of  the  student's  ability  to  perform  accurate  work.  Four  cutting  planes,  C  G,  C  B,  C  P,  and  C  E, 
are  shown  on  the  front  view,  and  the  curves  produced  by  them  are  to  be  drawn  on  the  top  view, 
they  being  projected  without  the  use  of  elements  of  the  cone.  The  revolved  section  of  each 
of  the  curves  is  next  to  be  found,  and  finally,  they  are  to  be  tested  by  the  following  methods : 
The  curve  of  the  ellipse  having  been  obtained,  find  its  axes  and  foci,  and  eight  points  by  the  first 
method  [page  44],  Test  the  curve  by  trammels  also.  The  parabola  and  hyperbola  are  to  be 
tested  by  the  second  methods  only.  In  testing  the  hyperbola,  it  should  be  noted  that  the  vertex 
of  the  conjugate  hyperbola  is  as  much  above  the  vertex  of  the  cone  as  the  vertex  of  the  given 
hyperbola  is  below  it.  Observe  that  the  cutting  plane  for  the  parabola,  C  F,  is  drawn  parallel 
to  an  element  of  the  cone. 


76  THE   INTERSECTION    OF    SURFACES. 


VI. 

THE    INTERSECTION    OF    SURFACES. 
PLATE  20, 

THREE  views  of  two  intersecting  cylinders  are  shown  in  Fig.  1,  and  it  is  required  to  find  their 
curve  of  intersection,  and  develop  the  cylinders.  On  the  side  view,  assume  an  element  of  the 
small  cylinder,  as  A8  Bs,  and  project  this  into  the  front  and  top  views.  AT  BT  is  seen  to  pierce 
the  large  cylinder  at  the  point  CT  BT,  which  is  the  top  view  of  an  element  of  this  cylinder.  Pro- 
jecting this  element,  we  shall  obtain  the  front  view  of  two  intersecting  elements,  CF  BF  and  AF  BF. 
This  point  of  intersection  will  therefore  be  common  to  both  cylinders,  and  hence  a  point  in  the 
required  curve  of  intersection.  In  like  manner  obtain  a  sufficient  number  of  points  to  determine 
the  curve.  For  convenience  in  the  development  of  the  small  cylinder,  it  is  desirable  to  have  its 
elements  equidistant. 

In  all  problems  relating  to  the  development  of  surfaces,  it  is  of  first  importance  to  determine 
those  points  of  the  curve,  known  as  limiting  points,  at  which  the  direction  of  curvature  changes, 
or  points  of  tangency  occur.  These  define  the  general  character  of  the  curve,  and  often  enable  it 
to  be  drawn  by  the  finding  of  a  less  number  of  points.  1F,  2P,  3F,  4F,  5F,  and  6F  are  limiting  points 
of  this  curve. 

In  developing  the  small  cylinder,  Fig.  2,  open  it  on  the  element  L  4,  which  will  make  it 
symmetrical  with  respect  to  the  centre.  The  method  of  developing  does  not  differ  from  Problem 
26,  and  the  length  of  the  elements  may  be  taken  from  the  front  or  top  views. 


THE    INTERSECTION    OF    SURFACES.  77 

The  development  of  the  large  cylinder,  Fig.  3,  will  be  a  rectangle  pierced  by  a  hole  which  is 
symmetrical  with  respect  to  a  horizontal  centre  line  only.  Having  obtained  the  development  of 
the  cylinder,  by  opening  it  on  G  H,  the  element  D  E  will  be  drawn  in  the  centre  of  the  surface, 
and  from  this  the  other  elements  obtained.  On  D  E  lay  off  the  points  3  and  5,  equidistant  from 
the  centre  line,  and  equal  to  that  portion  of  the  element  cut  by  the  small  cylinder,  as  shown  on  the 
front  view.  The  distance  between  any  of  these  elements,  as  D  E  and  C  N,  may  be  found  by 
measuring  the  circular  arc  DT  2T  CT,  which  should  then  be  laid  off  to  the  left  of  E  D,  and  through 
this  point,  C  N  may  be  drawn.  Having  the  position  of  an  element,  immediately  lay  off  the  amount 
cut  out  by  the  second  cylinder,  as  seen  on  the  front  view. 

PROBLEMS. 

PLATB  19. 

PROBLEM  29. —  This  differs  from  the  illustration  in  Plate  19  only  in  that  .the  axes  of  the 
cylinders  intersect. 

PROBLEM  30. —  Obtain  the  limiting  points  first.  Use  great  care  in  determining  the  curve,  as 
the  accuracy  of  the  development  is  entirely  dependent  on  it. 

PROBLEM  31,  plate  21. —  The  axes  of  these  cylinders  intersect,  but  are  not  at  right-angles  as 
in  the  preceding  case.  The  side  view  not  being  available  for  spacing  the  elements,  the  following 
method  may  be  pursued :  Draw  any  element  of  the  small  cylinder  AF  BF,  and  let  X  Y  be  a  circle 
of  the  cylinder  made  by  a  cutting  plane  perpendicular  to  the  axis.  If  this  circle  be  revolved 
about  X  Y,  parallel  to  the  front  plane,  the  point  CF,  which  is  a  point  of  both  the  element  and 
the  circle,  will  be  revolved  to  C',  and  CF  C'  will  be  the  distance  of  the  element  from  the  vertical 
plane  passing  through  the  axis.  This  will  determine  the  distance  between  the  centre  line,  and 


78  THE  INTERSECTION  OF  SURFACES. 

the  element  in  the  top  view,  the  latter  piercing  an  element  of  the  large  cylinder  at  AT.  Since 
there  is  a  second  element,  EF  DF,  of  the  small  cylinder,  lying  in  the  same  vertical  plane  as  the 
first  element,  it  will  intersect  the  same  element  of  the  large  cylinder  at  the  point  DF.  Thus  find 
any  number  of  points.  As  in  the  preceding  cases,  it  will  be  convenient  to  have  the  elements  of 
the  small  cylinder  spaced  equally,  and  this  may  be  done  by  spacing  them  on  the  revolved  position 
of  the  circle  X  Y,  and  through  these  points  drawing  the  elements.  In  developing  the  small 
cylinder,  it  will  be  found  necessary  to  assume  some  line  of  the  surface  which  on  being  developed 
will  be  a  straight  line,  for  in  this  problem,  the  development  of  the  cylinder  ends  will  be  curves. 
Such  a  line  will  lie  in  a  plane  perpendicular  to  the  axis,  and  the  section  made  by  the  plane  is 
called  a  right  section.  The  line  X  Y  fulfils  this  condition,  and  may  be  used  as  a  base  line  for 
obtaining  the  length  of  the  elements  in  the  development,  and  on  it  the  distances  between  the 
elements  may  be  measured. 

USE    OF    AUXILIARY    PLANES. 

PLATE  22. 

In  determining  the  curve  of  intersection  between  two  surfaces,  it  is  customary  to  use  a  system 
of  planes  which  shall  cut  either  circles  or  straight  lines  from  these  surfaces.  The  intersection  of 
these  lines  will  be  points  of  the  curve.  Let  it  be  required  to  find  the  intersection  of  cylinders 
1  and  2,  Fig.  1.  If  we  imagine  them  cut  by  a  plane  V  W,  parallel  to  the  axes,  the  appearance 
will  be  as  in  Fig.  3.  Two  elements  will  have  been  cut  from  each  cylinder,  and  since  they  lie  in 
the  same  plane,  their  points  of  intersection  will  be  common  to  both  cylinders,  and  therefore  in 
the  required  curve.  Again,  if  we  should  employ  a  plane  tangent  to  cylinder  2,  it  would  cut  that 
cylinder  in  a  line,  and  the  other,  in  two  elements,  as  in  Fig.  2.  These  last  points,  7  and  8,  would 
be  the  limiting  points  of  the  curve,  and,  ordinarily,  the  first  to  be  determined. 


THE  INTERSECTION  OF  SURFACES.  79 

To  obtain  the  elements  cut  by  these  planes,  proceed  as  follows :  Revolve  the  bases  of  the 
cylinders,  as  in  Problem  16,  and  having  assumed  a  cutting  plane,  as  V  W,  shown  on  the  top 
view,  lay  off  on  the  revolved  bases,  L  M  and  N  0,  equal  to  the  distance  of  the  cutting  plane  from 
the  plane  of  the  axes.  Through  the  points  M  and  0,  draw  parallels  to  the  bases,  and  these  will 
indicate  the  amount  cut  from  each  cylinder.  Next,  revolve  the  points  B,  D,  G,  and  K,  back 
into  the  bases,  and  through  them  draw  the  elements.  Their  intersection  at  3, 4, 5,  and  6  will  be  the 
four  points  determined  by  the  plane  V  W.  In  like  manner  the  points  7  and  8  may  be  found 
by  using  a  plane  tangent  to  the  small  cylinder.  The  points  in  the  top  view  are  obtained  by 
projecting  them  from  the  front  view  on  to  the  plane  in  which  they  lie.  Thus,  the  point  7  was 
found  by  means  of  the  cutting  plane  X  Y,  and  its  top  view  7T  must  therefore  lie  on  that  line. 

PROBLEMS. 

PLATE  21. 

PROBLEM  32. —  Having  obtained  front  and  top  views,  proceed  with  the  construction  of  the 
curve  of  intersection  by  first  using  a  tangent  plane  to  define  the  limits  of  the  curves.  In 
developing  the  large  cylinder,  which  alone  is  required,  open  it  on  an  element  not  cut  by  the 
second  cylinder. 

PROBLEM  33. —  It  is  required  to  find  the  intersection  of  a  cylinder  and  prism  without  using  a 
side  view.  Use  cutting  planes  parallel  with  the  vertical  face  of  the  prism.  The  base  of  the 
prism  in  the  front  view  will  have  to  be  revolved  in  order  to  complete  the  top  view  and  enable 
the  intersection  of  the  cutting  planes  and  prism  to  be  determined.  Both  surfaces  are  to  be 
developed.  A  thorough  understanding  of  previous  examples  should  enable  this  and  the  following 
problems  to  be  performed  with  little  instruction. 


80  THE    INTERSECTION    OF    SURFACES. 

PROBLEM  34. —  In  all  cases  of  intersection  between  prism  and  prism  it  is  only  necessary  to  find 
the  point  of  intersection  of  each  edge  of  both  prisms  with  a  face  of  the  other  prism.  To  avoid 
confusion,  these  should  be  taken  consecutively  as  follows :  Edge  A  B  of  the  triangular  prism  with 
face  A  C  of  the  hexagonal ;  edges  C  and  D  of  the  hexagonal  prism  with  face  E  F  B  A.  of  the 
triangular;  then  E  F  with  face  D  G, etc.  Much  care  will  be  required  in  developing  these  surfaces. 

PROBLEM  35. —  The  lines  of  intersection  on  the  top  view  are  to  be  completed,  and  the  front 
view,  with  its  lines  of  intersection,  are  required.  Develop  the  prism  only. 

PROBLEM  36. —  The  intersecting  prisms  being  at  an  angle  other  than  90°  makes  this  problem 
more  difficult  than  the  preceding.  To  obtain  the  point  C,it  will  be  necessary  to  first  find  the  top 
view  of  the  line  of  intersection  of  a  plane  coinciding  with  the  upper  face  of  the  prism  A,  with  the 
face  B,  of  the  second  prism.  The  intersection  of  this  line  with  the  upper  edge  D  0,  will  deter- 
mine the  required  point.  Develop  the  prism  A. 


SCREW-THREADS    AND    BOLT-HEADS.  81 

VII. 

SCREW-THREADS    AND    BOLT-HEADS. 

SPIRALS. 

PLATE  23. 

The  Spiral  is  a  curve  generated  by  a  point  moving  in  a  plane  about  a  centre,  from  which  its 
distance  is  continually  increasing. 

Imagine  a  right  line,  A  B,  Fig.  1,  free  to  revolve  in  the  plane  of  the  paper  about  one  of  its 
extremities,  A,  as  an  axis.  Also  conceive  a  point  as  free  to  move  on  this  line.  Three  classes  of 
lines  may  now  be  derived  in  the  following  manner  :  If  the  line  be  stationary  and  the  point  move, 
a  straight  line  will  be  generated ;  if  the  point  be  stationary  and  the  line  revolve,  a  circle  will  be 
generated  ;  if  both  the  point  and  line  move,  a  curve  known  as  a  SPIRAL  will  be  generated.  The 
character  of  this  curve  varies  with  the  relative  motions  of  point  and  line.  If,  during  one  uniform 
revolution  of  the  line,  the  motion  of  the  point  be  uniform,  the  SPIRAL  OF  ARCHIMEDES,  or  EQUABLE 
SPIRAL,  will  be  generated.  The  line  A  B,  Fig.  1,  is  the  Radius  Vector,  and  the  radial  distance 
traversed  by  the  point  during  one  revolution  is  called  the  PITCH.  By  varying  the  relative  motions 
of  the  line  and  point,  all  classes  of  spirals  may  be  generated. 

Fig.  1  illustrates  the  equable  spiral,  and  the  method  for  drawing  the  same.  Twelve  succes- 
sive positions  of  the  radius  vector  being  shown  by  A  C,  A  D,  A  E,  etc.,  the  distance  of  the  point 
from  the  centre  will  be  increased  by  one-twelfth  of  the  pitch  during  each  twelfth  of  a  revolution 

6 


82  SCREW-THREADS. 

of  the  radius  vector.     Having  found  a  sufficient  number  of  points,  lightly   sketch  the  curve 
freehand. 

INVOLUTES.  — These  curves  are  a  class  of  spirals  which  may  be  generated  by  the  unwinding 
of  a  perfectly  flexible,  but  inextensible  cord,  from  a  polygon  of  any  number  of  sides,  the  names 
of  the  involutes  being  derived  from  the  polygons  which  determine  their  form.  The  curves  con- 
sist of  circular  arcs,  having  for  centres  the  vertices  of  the  polygons,  and  radii  increasing  by  an 
amount  equal  to  the  length  of  the  sides.  Fig.  2  is  the  involute  of  a  square,  drawn  by  describing 
the  quadrant  4-5  from  centre  1,  with  radius  1-4,  then  the  quadrant  5-6  from  centre  2,  with  radius 
5-2,  and  so  on,  until  the  desired  length  of  curve  shall  have  been  drawn.  If  two  opposite  sides  of 
the  square  become  infinitely  small,  the  result  will  be  a  right  line,  the  involute  of  which  is  shown 
in  Fig.  8.  Again,  if  the  number  of  sides  of  the  polygon  become  infinite,  we  shall  obtain  the 
involute  of  a  circle. 

THE    HELIX. 

PLATE  23. 

If  the  line  on  which  the  generating  point  is  supposed  to  move,  be  made  to  revolve  about  an 
axis  with  which  it  makes  an  angle  of  less  than  90°,  thus  generating  a  cylinder  or  cone,  a  class  of 
curves  known  as  HELICES  will  be  described.  If  the  line  A  B,  Fig.  4,  be  parallel  to  the  axis 
about  which  it  revolves,  and  the  generating  point  move  on  this  line,  three  classes  of  lines  may  be 
described,  as  in  the  case  of  spirals.  A  circle  will  be  generated  when  the  point  is  stationary  and 
the  line  revolves ;  a  right  line  is  generated  when  the  line  is  stationary  and  the  point  moves  on  the 
line ;  and  since  the  circle  and  right  line  do  not  lie  in  the  same  plane,  the  result  will  be  a  HELIX 
when  these  motions  take  place  simultaneously.  The  distance  traversed  by  the  generating  point 
on  the  line  A  B,  during  one  revolution,  is  called  the  PITCH.  The  motion  of  line  and  point  being, 


SCREW-THREADS.  83 

in  general,  uniform,  the  curve  is  described  as  follows:  Having  assumed  any  desired  number  of 
equidistant  positions  of  the  generating  element,  A  B,  as  1,  2, 3, 4,  etc.,  the  pitch  should  be  divided 
into  the  same  number  of  parts,  as  shown  by  the  horizontal  lines  drawn  through  1',  2',  3',  4',  etc. 
When  the  element  shall  have  moved  through  one  of  its  divisions  to  the  position  1,  which  in  this 
case  is  one-twelfth  of  a  revolution,  the  generating  point  will  have  moved  on  the  generating  line 
through  one-twelfth  of  the  pitch,  and  the  point  K  will  be  determined.  Also  when  the  element 
has  made  one-quarter  of  a  revolution,  the  point  will  have  traversed  one-quarter  of  the  pitch,  and 
be  at  C.  As  the  rate  of  curvature  is  most  rapid  at  the  points  of  tangency,  A,  D,  and  B,it  is 
desirable  to  obtain  a  greater  number  of  points  by  subdivision,  as  shown  in  the  figure. 

Fig.  5  illustrates  the  development  of  that  portion  of  the  cylinder  lying  below  the  helix,  in 
which  it  will  be  seen  that  the  development  of  the  curve  is  a  right  line.  If  a  triangle  be  con- 
structed, having  for  its  base  the  circumference  of  the  cylinder,  and  for  its  perpendicular  the  pitch 
of  the  helix,  the  hypothenuse  will  be  the  development  of  the  helix. 

PROBLEMS. 

PLATE  24. 

PROBLEM  37.  —  In  drawing  the  equable  spiral,  describe  more  than  one  revolution  of  the  radius 
vector,  and  find  at  least  twenty-four  points  in  each  revolution.  Sketch  the  curve  in  pencil,  and 
ink  all  save  the  outer  portion  by  means  of  compasses. 

The  involutes,  being  composed  of  circular  arcs,  will  be  described  entirely  by  means  of 
compasses. 

PROBLEM  38.  —  In  the  first  example  observe  that  the  pitch  of  the  helix  is  but  2  inches, 
while  the  length  of  the  cylinder  is  2|. 


84  SCREW-THREADS. 

In  Fig.  4,  Plate  23,  the  helix  is  right-handed  and  single,  but  if  it  were  required  to  be  double, 
it  would  only  be  necessary  to  draw  a  parallel  curve  beginning  at  the  point  6',  opposite  I). 

If  the  generating  element  shall  describe  a  cone  instead  of  a  cylinder,  the  curve  generated  will 
be  a  conical  helix,  the  top  view  of  which  is  an  equable  spiral.  The  pitch  is  measured  parallel  to 
the  axis,  as  in  preceding  cases. 

In  inking  helices,  always  use  the  bow  pen  for  a  small  part  of  the  curve  at  the  points  of  tan- 
gency,  and  since  the  curve  is  symmetrical  on  either  side  of  the  axis,  be  particular  to  use  the  same 
part  of  the  scroll  for  the  inking  of  each  side,  reversing  the  scroll  in  each  case. 

V   AND    SQUARE    SCREW-THREADS. 

PLATES  25  and  26. 

In  order  to  make  the  drawing  of  a  screw-thread,  it  is  necessary  to  know  :  The  character  of 
the  thread,  whether  V  (as  in  Plate  25),  or  square  (as  in  Plate  26)  ;  whether  single  or  double; 
the  diameter,  and  the  pitch.  If  the  angle  of  the  V  thread  is  other  than  60°,  that  also  must 
be  known. 

If  the  thread  to  be  drawn  has  a  V  section,  as  in  Fig.  1,  Plate  25,  and  the  diameter  and  pitch 
are  given,  begin  by  drawing  the  section,  as  shown  in  dotted  lines,  throughout  the  length  of  the 
screw.  Next  construct  a  templet  as  follows:  Draw  on  a  rather  light  piece  of  card-board,  two 
helices  of  the  same  pitch  as  the  required  thread,  one  being  of  a  diameter  equal  to  the  outside  of 
the  thread,  and  the  other  equal  to  the  diameter  at  the  root  of  the  thread.  These  may  be  drawn 
separately,  or  together,  as  in  Fig.  2,  and  are  to  be  carefully  cut  out  so  as  to  be  used  as  a  pattern 
in  pencilling  the  curves.  Having  drawn  the  helices,  mark  the  points  of  tangency  B,  G,  and  A,  F, 
as  well  as  the  centres  L  and  K,  in  the  manner  indicated.  Also  repeat  those  marks  on  the  oppo- 


SCREW-THREADS.  85 

site  side  after  cutting  out.  This  will  enable  the  templet  to  be  used  either  side  up,  and  be  readily 
set  to  the  drawing-.  Observe  that  the  curve  is  not  to  be  cut  off  abruptly  at  its  termination,  but 
continued  a  little  beyond,  so  that  in  tracing  the  outline,  the  pencil  point  may  not  injure  the 
extreme  point  of  the  templet  curve.  This  may  now  be  used  for  the  drawing  of  all  the  helices 
on  this  screw,  as  from  A  to  F,  B  to  G,  C  to  H,  etc.,  Fig.  1.  If  the  pitch  were  small  in  propor- 
tion to  the  diameter,  the  drawing  of  the  screw  might  now  be  considered  finished  ;  but  in  order  to 
correctly  illustrate  the  projection  of  a  V  thread,  we  must  consider  the  character  of  the  surface, 
and  apply  a  correction  to  the  drawing.  Therefore,  imagine  the  line  A  13  to  revolve  about  the 
axis  of  the  screw,  and  at  the  same  time  move  in  the  direction  of  the  axis  so  as  to  generate  the 
upper  half  of  the  surface  of  the  screw.  Every  point  of  the  line  will  describe  a  helix,  the  diame- 
ters differing,  but  the  pitch  remaining  constant.  The  helices  generated  by  the  extremities  of  the 
line  have  already  been  drawn,  and  the  curves  123  and  456,  described  by  two  other  points.  1 
and  4,  are  shown  by  the  fine  dotted  lines.  The  curved  line  M  5  2  N,  drawn  tangent  to  these 
helices,  will  be  the  visible  outline  of  the  surface,  instead  of  the  dotted  line,  which  is  concealed. 
As  the  labor  of  describing  these  helices  would  be  great,  it  is  customary  to  draw  the  outlines  of 
the  screw  as  follows  :  Having  described  the  helices  A  F,  B  G,  etc.,  reverse  the  templet  and  draw 
a  small  portion  of  the  continuation  of  the  helix  on  the  opposite  side  of  the  screw,  as  at  B  P  and 
C  0.  Then  tangent  to  the  two  helices  draw  the  line  M  N,  and  in  the  other  direction  the  tangent 
0  P,  a  portion  of  which  is  invisible.  It  will  be  observed  that  more  than  half  the  outer,  and  less 
than  half  the  inner  helix,  is  visible  in  each  case. 

Fig.  3  represents  the  nut  for  this  thread,  but  as  this  is  shown  in  section,  the  helices  of  the 
opposite  side  alone  are  visible.  The  outer  helices  are  concealed  at  their  extremities. 

Plate  26  represents  a  single,  and  double,  square  thread  and  nut.  These  are  described,  as  in 
the  preceding  case,  by  the  making  of  a  templet  and  afterwards  using  it  to  trace  the  curves,  the 


86  SCREW-THREADS. 

section  of  the  thread  being  first  completed.  The  square  section  of  the  thread  is  equal  to  one-half 
the  pitch  in  the  single,  and  one-quarter  in  the  double  thread.  The  character  of  the  latter  will  be 
more  fully  explained  in  Plate  27. 

PROBLEM  39,  Plate  24.  —  Having  drawn  the  section  of  the  screw  and  nut,  prepare  the  templet, 
and  since  the  diameter  and  pitch  are  the  same  in  both  examples,  it  may  be  used  for  the  drawing 
of  all  the  helices. 

In  the  inking  of  the  curves  follow  the  directions  iriven  for  Problem  38. 


CONVENTIONAL,    THREADS. 

PLATE  27. 

In  order  to  facilitate  the  drawing  of  screw-threads,  it  is  customary  to  omit  the  drawing  of  the 
helix,  substituting  therefor  a  right  line,  as  in  Fig.  1.  In  most  cases,  however,  even  this  would 
involve  too  much  labor,  as  well  as  complication  on  the  drawing,  and  the  Vs  are  likewise  omitted, 
the  representation  shown  in  Fig.  6  being  adopted.  When  this  is  done,  no  pains  are  taken  to 
make  the  screw  of  the  required  pitch,  and  the  spaces  between  the  fine  lines,  which  represent  the 
outer  helices,  are  estimated  by  the  eye,  as  in  section  lining.  But  it  is  imperative,  for  the  proper 
representation  of  a  single  thread,  that  the  point  C  be  over  the  middle  of  the  space  B  D.  Having 
drawn  the  line  A  B,  make  the  space  A  C  double  that  of  E  B,  and  draw  parallels.  Afterwards, 
draw  the  heavy  lines  to  represent  the  root  of  the  thread.  As  it  is  difficult  to  draw  these  of  equal 
length  without  the  aid  of  special  lines,  draughtsmen  have  come  to  adopt  the  method  illustrated 
Fig.  7r  which  is  the  more  practical  representation,  and  the  one  usually  employed. 

THE  DOUBLE  THREAD. — A^  the  use  and  character  of  the  double  thread  is  generally  misunder- 
stood, Figures  1,  2,  and  3  have  been  drawn  to  more  clearly  explain  this  problem.  Fig.  1  illus- 


SCREW-THREADS.  87 

trates  a  screw,  the  diameter  and  pitch  of  which  are  supposed  to  have  been  given  ;  but  as  the  pitch 
is  excessive,  for  a  screw  of  this  diameter,  the  diameter  at  the  root  of  the  thread  is  small,  and  the 
screw  proportionally  weak.  The  only  way  to  strengthen  the  thread  at  this  point,  without  chang- 
ing the  angle  of  the  V's,  is  by  making  a  screw  with  the  V's  partly  filled  at  the  root,  as  shown  by 
the  dotted  line  in  Fig.  1,  and  the  complete  screw  in  Fig.  2,  thus  increasing  the  diameter  at  the 
root.  While  overcoming  one  weakness  we  have  introduced  a  second,  by  lessening  the  section  of 
the  thread,  so  that  with  a  nut  of  a  given  length,  the  tendency  of  the  thread  to  be  stripped  from 
the  body  is  doubled.  This  last  difficulty  may  be  overcome  by  supposing  an  intermediate  thread 
wound  between  the  present  threads,  as  shown  in  Fig.  2  by  the  dotted  lines,  and  also  by  the  com- 
pleted drawing  of  Fig.  3,  which  is  a  representation  of  a  double  thread  having  the  same  diameter 
and  pitch  as  the  single  thread  of  Fig.  1,  but  of  increased  strength.  It  must  be  noted  that  the  full 
and  dotted  threads  of  Fig.  2  are  entirely  independent  of  each  other,  and  that  the  point  C  of  one, 
is  diametrically  opposite  a  point  B  in  the  parallel  thread.  This  must  be  carefully  observed  in  the 
practical  representation  of  a  double  thread,  as  shown  in  Fig.  8. 

Fig.  9  represents  a  left-hand  single  thread. 

The  conventional  square  thread  is  illustrated  in  Fig.  5,  and  this  may  also  be  simplified  by 
omitting  the  lines  representing  the  root  of  the  thread. 

A  section  of  a  U.  S.  STANDARD  V  THREAD  is  shown  in  Fig.  4,  and  the  proportions  for  the  same 
are  as  follows  :  — 

D  =  Diameter  of  bolt. 

P  =  Pitch  of  thread  =  0.24  VD  -f  0.625  —  0.175. 

N  =  Number  of  threads  per  inch  =  ^. 

S  =  Depth  of  thread  =  0.65  P.      The  angle  of  V's  is  always  60°. 

PROBLEM  40,  Plate  24.    The  first  eight  examples  are  to  be  drawn  by  the  conventional  methods 


BOLTS. 


illustrated  in  Figs.  1,  3,  and  5,  of  Plate  27,  care  being  used  in  observing  all  the  conditions  pre- 
scribed. The  last  four  examples  are  to  be  drawn  by  the  methods  illustrated  in  Figs.  6,  7,  8,  and 
9,  of  the  same  Plate.  Make  the  pitch  for  single  threads  about  equal  to  that  shown  in  Fig.  1, 
Plate  29,  estimating  the  spaces  by  the  eye. 

BOLTS. 

SPHERE  AND  CUTTING  PLANES. 

PLATE  28. 

If  a  sphere  be  cut  by  six  planes  equidistant  from,  and  parallel  to,  a  vertical  axis,  a  represen- 
tation will  be  obtained  similar  to  that  of  a  hexagonal  bolt-head  or  nut,  of  which  a  careful  study 
should  be  made. 

Fig.  1  represents  three  views  of  a  sphere  cut  by  planes  V  W,  XY,  etc.,  as  prescribed  above. 
Any  cutting  plane  will  intersect  the  sphere  in  a  circle,  and  if  the  cutting  plane  is  parallel  to  the 
plane  of  projection,  this  intersection  will  appear  as  a  circle,  but  otherwise  as  an  ellipse  or 
straight  line. 

The  plane  VW  intersects  the  sphere  in  a  circle  the  diameter  of  which  is  ET  FT,  and  slioun 
on  the  front  plane  by  EF  AF  BF  FF.  The  plane  X  Y  also  intersects  the  sphere  in  a  circle  of  equal 
diameter,  being  at  the  same  distance  from  the  axis;  but  this  plane  being  inclined  to  the  front 
plane,  the  circle  will  appear  as  an  ellipse,  having  a  major  axis  GF  HF,  equal  to  the  diameter  of 
the  circle,  and  the  minor  axis  KF  LF,  equal  to  the  foreshortened  diameter  projected  from  the  top 
view.  The  planes  V  W  and  X  Y  intersect  in  the  line  BF  0F,  thus  cutting  off  a  portion  of  the 
circle  and  ellipse.  Similarly,  the  remaining  curve  of  intersection,  DF  AF,  may  be  found,  and 
the  side  view  also  obtained. 


BOLTS. 

Points  in  the  curve  may  also  be  found  by  revolving  the  plane  XY  parallel  to  the  top  plane, 
thus  obtaining  the  height  of  any  point,  as  N,  which  may  then  be  projected  into  the  other  views. 

Fig.  2  is  a  similar  example,  in  which  four  cutting  planes  are  used,  thus  making  it  similar  to  a 
square-headed  bolt  or  nut. 

It  is  desirable  that  the  details  of  these  figures  be  carefully  worked  out  by  the  student,  treating 
this  plate  as  a  problem. 

HEXAGONAL   HEADS   AND   NUTS. 

PLATE  29. 

Since  we  may  regard  the  upper  half  of  front  and  side  views  of  Fig.  1,  Plate  28,  as  a  true  repre- 
sentation of  a  hexagonal  bolt-head,  save  that  the  proportion  is  not  good,  it  is  important  to 
consider  the  salient  points  of  this  representation  that  they  may  be  applied  to  the  drawing  of  bolt- 
heads  and  nuts. 

First :  Three  faces  of  the  head  are  seen  when  it  is  shown  "  across  corners,"  as  in  the  front 
view,  and  one  of  these  is  double  the  width  of  the  other  two. 

Second :  The  circular  arc  AF  MF  BF  is  concentric  with  the  circle  of  the  sphere,  and  the  points 
AF  and  BF  are  determined  from  the  height  of  CF,  a  point  of  the  intersection  of  the  two  planes,  or 
faces  of  the  head,  and  the  great  circle  of  the  sphere. 

Third :  The  major  axes  of  the  ellipses  and  the  diameters  of  the  circles,  made  by  the  cutting 
planes,  are  equal,  hence  points  GF  and  MF  are  at  the  same  height. 

Fourth :  Two  equal  faces  are  seen  when  the  head  is  shown  "across  flats,"  as  in  the  side  view. 

Fifth  :  The  points  Ms  and  Gs  are  of  equal  height,  as  in  the  front  view,  and  Bs  must  be 
obtained  by  projection  from  the  front  view,  or  in  the  same  manner  as  CF. 


90  BOLTS. 

The  proportion  for  a  bolt-head  or  nut  is  shown  in  Fig.  1,  Plate  29,  which  is  an  illustration  of 
a  U.  S.  Standard  bolt.  Fig.  5  is  also  a  standard  bolt,  but  with  the  ends  chamfered  instead  of 
rounded.  Either  of  these  types  may  be  used,  the  former  being  employed  to  represent  a  finished 
head  only,  and  the  latter  for  rough  and  finished  heads. 

Having  considered  both  the  representation' and  proportion  of  the  head,  we  may  proceed  with 
the  practical  drawing  of  bolt-heads  and  nuts.  Suppose  it  is  required  to  draw  a  rounded  head 
"across  corners,"  as  shown  in  Fig.  2, Plate  29.  Having  drawn  the  centre  line,  underside  of  head, 
and  diameter,  as  indicated  by  lines  1,  2,  3,  and  4,  figure  the  short  diameter  of  the  hexagon,  or 
distance  "  across  flats."  According  to  the  proportion  given,  this  is  equal  to  the  diameter  of  bolt, 
pins  one-half  the  diameter,  plus  one-eighth  of  an  inch.  Lay  off  E  F  equal  to  one-half  this  amount, 
and  draw  the  perpendicular  F  G  and  the  30°  line  E  G,  then  will  the  triangle  E  F  G  represent 
one-twelfth  of  the  top  view  of  the  head,  and  E  G  will  be  equal  to  half  the  long  diameter  required. 
Lay  off  this  distance  on  either  side  of  E,  thus  determining  lines  8  and  9.  Draw  10  and  11, 
remembering  that  these  lines  equally  divide  the  spaces  between  1  and  8,  and  1  and  9,  which 
spaces  also  equal  twice  F  G.  Next  determine  the  thickness  of  head,  and  with  a  radius  1  equal  to 
twice  the  diameter  of  the  bolt,  describe  arc  12,  which  determines  points  D,  C,  A,  and  B.  From 
the  same  centre  describe  arc  14.  Arcs  15  and  16  should  be  drawn  as  circular  arcs,  their  height 
being  determined  from  14. 

If  it  is  required  to  represent  a  bolt-head  "  across  flats,"  as  in  Fig.  3,  proceed  as  before,  deter- 
mining the  short  diameter,  and  drawing  5  and  6.  Next  figure  the  thickness  of  head  and  describe 
arc  7  ;  this  will  determine  E,  and  the  height  of  arcs  12  and  13.  Although  these  arcs  are  elliptical 
in  theory,  they  should  always  be  described  as  circular.  To  determine  the  point  A,  we  must  find 
the  long  diameter,  as  in  the  previous  case,  and  so  obtain  line  10,  the  intersection  of  which  with 
7,  will  be  at  D,  equal  in  height  to  A,  B,  and  C.  Finally  draw  arcs  12  and  13. 

1  There  is  no  standard  for  this  radius,  but  2  D  is  recommended  as  being  a  convenient  radius  for  draughtsmen. 


BOLTS.  91 

Fig.  4  illustrates  a  rounded  nut  "  across  corners,"  the  order  for  the  drawing  of  the  lines  being 
indicated  by  the  figures.  It  must  be  observed,  however,  that  the  thickness  of  a  nut  differs  from 
that  of  the  head,  being  equal  to  the  diameter  of  the  bolt ;  also  that  since  the  nut  is  pierced  by  a 
hole,  the  top  will  appear  flat,  and  the  arc  13  must,  therefore,  be  struck  from  K,  and  with  radius 
equal  to  2  D. 

If  we  substitute  a  cone  for  the  sphere,  in  Plate  28,  cutting  it  by  the  six  vertical  planes  par- 
allel to  the  axis,  and  then  pass  a  seventh  plane  perpendicular  to  the  axis,  and  tangent  to  the 
curves  of  intersection,  a  representation  will  be  obtained  similar  to  that  of  the  head  and  nut  in 
Fig.  5,  Plate  29.  This  is  called  a  chamfered  bolt,  and  the  curve  of  intersection  is  an  hyperbola; 
but  since  these  curves  approximate  circular  arcs,  the  following  concise  method  may  be  employed. 
If  it  is  required  to  draw  a  chamfered  head  or  nut  across  corners,  construct  a  hexagonal  prism  of 
dimensions  required  for  a  standard  bolt.  From  the  centre  line,  with  radius  equal  to  diameter  of 
bolt,  describe  the  arc  A  B,  tangent  to  end  of  nut,  and  from  the  same  centre  draw  arcs  D  E,  and 
C  F,  points  D  and(C  being  determined  by  A  and  B.  Finally  draw  arcs  D  A  and  B  C,  also  tangent 
to  E  F.  The  method  for  drawing  the  chamfered  head  or  nut  "  across  flats  "  is  apparent  from  the 
illustration  of  the  bolt-head  in  Fig.  5. 

PROBLEM  41,  Plate  29.  —  The  diameters  are  given,  and  the  sketch  shows  the  character  of  the 
bolt,  whether  rounded  or  chamfered.  Use  care  in  observing  every  detail,  and  see  that  the  dimen- 
sions are  standard.  It  is  better  to  draw  first  the  rounded  heads  and  nut,  as  shown  at  the  left  of 
the  problem. 

For  the  further  consideration  of  this  subject  the  student  is  referred  to  the  chapter  on  Bolts 
and  Screws,  in  Machine  Drawing  of  this  series. 


92 


ISOMETRIC    AND    OBLIQUE    PROJECTION. 


VIII. 

ISOMETRIC    AND    OBLIQUE    PROJECTION. 

IT  is  frequently  necessary  to  produce  a  pictorial  effect  in  mechanical  drawings,  while  also 
preserving  the  relative  proportion  of  parts  so  that  the  drawings  may  be  made  to  scale,  and 
measurements  readily  taken  therefrom. 

Three  methods  of  accomplishing  this  are  as  follows  :  — 

First:  by  revolving  the  object  so  that  three  of  its  faces,  which  are  at 
right  angles  to  each  other,  shall  be  equally  inclined  to  a  plane  upon  which 
it  is  orthographically  projected.  This  is  called  the  Isometric  Projection  of 
the  object. 

Second  :  by  revolving  the  object  into  such  a  position,  relative  to  the  plane 
of  projection,  that  two  of  its  visible  faces  shall  be  foreshortened  double 
that  of  the  other  one,  and  projecting  as  before.  This  is  a  modification  of 
the  preceding. 

Third:    by  projecting  the  object  by  lines  oblique  to  the  plane  of  projection,  one 
face  of  the  object  being  parallel  to  the  plane.     This  system  is  known  as  Oblique 
Projection. 

It  will  be  observed  that  in  each  case  three  faces  of  the  object  are  visible,  and 
the  drawing  is  so  made  that  measurements  may  be  taken  from  either.  One  view 
is  thus  made  to  serve  for  two  or  more,  and  a  perspective  effect  is  also  obtained.  These  methods 


ISOMETRIC    PROJECTION.  93 

are  well  adapted  to  the  illustrating  of  much  architectural  work,  the  general  views  of  Patent-Office 
drawings,  the  sketching  of  machinery,  etc. 

ISOMETRIC    PROJECTION. 

PLATES  30,  31,  and  32. 

As  a  cube  best  illustrates  the  principles  of  isometric  projection,  conceive  one  as  having  been 
revolved  about  a  vertical  axis  until  the  diagonals  of  its  base  are  parallel  to  the  vertical  projecting 
planes.  Fig.  1  represents  the  side  view  of  the  cube  when  so  revolved.  Next,  incline  it  toward 
the  front  plane,  as  in  Fig.  2,  until  the  diagonal  E  C  is  horizontal,  and  draw  the  front  view  of  the 
same.  This  view  is  shown  by  Fig.  3,  which  illustrates  the  object  with  all  of  its  edges  equally 
foreshortened,  and  capable  of  being  measured  by  a  special  scale.  Furthermore,  it  will  be  seen 
that  all  of  the  edges  of  the  cube,  except  the  vertical,  make  angles  of  30°  with  the  horizontal. 

If  a  scale  of  equal  parts  were  constructed  for  the  measurement  of  Fig.  3,  its  units  would  be 
.81647  of  an  inch.  The  inconvenience  due  to  the  use  of  such  a  scale  has  caused  the  adoption  of 
the  full  scale  for  all  isometric  representations.  Fig.  6  illustrates  the  same  cube  drawn  to  the  full 
scale,  and  this  is  distinguished  from  the  preceding  representation  by  being  called  the  isometric 
drawing,  instead  of  the  isometric  projection,  of  the  object.  It  will  be  seen  that  this  representation 
is  larger  than  the  object ;  but,  it  is  more  easily  drawn,  and  serves  equally  well  for  an  illustration. 

The  lines  C  D?  C  B,  and  C  G  are  called  isometric  axes;  and  lines  parallel  to  them  are  known 
as  isometric  lines.  It  is  evident  that  only  isometric  lines  may  be  measured,  since  they  alone  are 
equally  foreshortened.  Thus,  the  isometric  of  the  diagonals  of  the  square,  A  C  and  D  B,  are  of 
unequal  length  although  in  the  original  we  know  them  to  be  equal.  Likewise,  it  is  not  possible 
to  directly  measure  the  angle  between  lines  on  an  isometric  drawing. 


94  ISOMETRIC    PROJECTION. 

To  make  the  isometric  drawing  of  a  cube,  Fig.  6.  —  From  the  point  C  draw  lines  C  B  and  C  D 
at  angles  of  30°,  and  equal  to  the  required  length  of  the  edges.  Draw  the  vertical  C  G,  of  same 
length.  As  each  edge  of  the  cube  is  parallel  to  one  or  the  other  of  these  isometric  lines,  the 
drawing  may  be  completed  as  shown.  In  shading  an  isometric  drawing,  it  is  customary  to  draw 
shade  lines  for  the  division  between  light  and  dark  surfaces,  the  direction  of  the  light  being 
that  of  the  diagonal  D  F.  This  makes  the  upper  and  left-hand  vertical  faces,  light  surfaces,  all 
others  being  dark. 

To  inscribe  a  circle  on  the  face  of  a  cube,  Figs.  5  and  6. —  Let  D'  C'  G'  H',  Fig.  5,  be  a  view  of  the 
face  on  which  the  circle  is  inscribed.  As  the  isometric  drawing  of  every  circle  is  an  ellipse,  it 
is  only  necessary  to  obtain  the  major  and  minor  axes  to  enable  the  curve  to  be  described  by  the 
method  of  trammels.  These  axes  lie  on  the  diagonals  D  G  and  H  C,  and  their  extremities  may 
be  determined  as  follows :  The  diagonals  being  non-isometric  lines,  the  distance  D  K  cannot 
be  measured,  and  the  point  K  must  be  determined  by  measurements  parallel  to  the  isometric 
axes,  as  P'  K'  and  0'  K',  which  equal  P  K  and  0  K.  Through  K,  an  extremity  of  the  major 
axis,  draw  K  N  parallel  to  D  C,  and  its  intersection  with  the  second  diagonal,  at  N,  will  deter- 
mine one  extremity  on  the  minor  axis.  Having  obtained  M  and  N,  as  indicated,  draw  the 
ellipse  by  the  method. of  trammels.  As  absolute  accuracy  is  not  always  a  requisite  of  isometric 
drawing,  the  labor  of  representing  ellipses  is  often  lessened  by  the  following  approximate  method. 
Bisect  the  edges  of  the  upper  face,  Fig.  6,  and  connect  these  points,  S,  T,  R,  V,  with  the  vertices 
A  and  C.  From  their  points  of  intersection,  Y  and  Z,  describe  arcs  R  S  and  T  V,  and  from 
centres  C  and  A,  describe  arcs  S  T  and  R  V. 

To  lay  off  any  required  angle,  Figs.  6  and  7. —  From  C',  on  C'  B',  Fig.  7,  angles  of 
15°,  30°,  45°,  60°,  and  75°  have  been  drawn,  and  it  is  required  to  construct  the  isometric  drawing 
of  similar  angles.  As  the  points  1',  2',  F',  etc.,  which  determine  the  required  angles,  lie  on 


ISOMETRIC    PROJECTION.  95 

isometric  lines,  it  is  only  necessary  to  transfer  them  to  the  isometric  drawing  of  the  face,  as  at 
1,  2,  F,  3,  etc.,  to  enable  the  required  line  to  be  drawn.  Points  in  the  circular  arc,  as  5  and  6, 
may  be  obtained  in  the  manner  shown  by  the  dotted  lines. 

To  obtain  the  isometric  drawing  of  an  object  similar  to  Fig.  4. —  The  rectangular  plinth  is  drawn 
in  like  manner  to  the  cube,  A  B,  A  D,  and  A  C  being  isometric  lines.  This  base  is  surmounted 
by  the  frustum  of  a  pyramid,  the  upper  base  of  which  is  a  square.  To  determine  E,  one  corner  of 
this  square,  lay  off  F  H  equal  to  the  perpendicular  distance  of  E  from  face  BAP,  and  G  F 
equal  to  its  distance  from  CAD;  at  their  intersection  erect  the  perpendicular  F  E  equal  to  the 
required  height  of  the  frustum  of  the  pyramid.  Next  construct  the  isometric  of  the  square. 
A  cylinder  of  height  K  L  rests  on  this  surface,  and  the  circles  of  its  bases  are  drawn  by  the 
approximate  method.  The  cylinder  is  surmounted  by  a  square  abacus,  the  vertical  faces  of 
which  are  pyramids  constructed  as  shown. 

When  a  figure  consists  of  lines  which  are  mostly  non-isometric,  it  may  be  referred  to  axes 
which  are  isometric,  as  in  Fig.  9,  Plate  31,  which  represents  the  isometric  drawing  of  a  regular 
pentagon.  D  E  being  chosen  as  an  isometric  line,  or  axis,  a  second  axis  Y  Y  is  drawn  isometri- 
cally  perpendicular  to  it.  Points  C  and  F  lie  on  this  axis,  and  their  position  is  readily  deter- 
mined. As  line  A  B,  drawn  through  C,  is  an  isometric  line,  points  A  and  B  may  be  located  at 
their  proper  distance  to  right  and  left  of  Y  Y. 

To  make  an  iso?netric  drawing  of  an  oblique  timber  framed  into  a  horizontal  timber,  Figs.  10 
and  1^ —  Fig.  10  illustrates  a  side  view  of  the  timbers ;  and  they  being  square,  it  supplies  the 
necessary  data  for  the  drawing.  After  having  drawn  the  lower  piece,  as  in  Fig.  11,  the  step  for 
the  oblique  timber  should  be  shown.  The  edges  of  this  cut  being  non-isometric,  they  must  be 
obtained  by  locating  points  C,  D,  and  E,  as  in  Fig.  11.  Suppose  the  required  pitch  of  the  oblique 
timber  to  be  two-thirds :  that  is,  two  vertical  units,  for  every  three  horizontal  units.  From  C, 


96  ISOMETRIC    PROJECTION. 

Fig.  11,  lay  off  on  A  B  any  three  units,  and  erect  a  perpendicular  equal  to  two  of  those  units. 
The  point  found  will  determine  the  pitch  of  the  oblique  timber.  Although  II'  L'  is  perpendicular 
to  H'  C',  Fig.  10,  it  is  not  possible  to  make  this  measurement  directly  on  the  isometric  drawing, 
as  it  is  anon-isometric  line  ;  but  if  a  perpendicular  H'  K'  be  drawn  from  H',  this  distance  may  be 
laid  off  from  H,  Fig.  11,  thereby  determining  a  point  K,  on  the  lower  side  of  the  timber,  through 
which  a  parallel  to  H  C  may  be  drawn.  Finally,  the  point  L  may  be  obtained  by  laying  off  M  N 
equal  to  M'  N',  and  erecting  a  perpendicular  to  intersect  lower  edge  of  timber. 

Plate  32  illustrates  the  application  of  the  foregoing  principles  to  the  representation  of  archi- 
tectural work.  The  figure  on  the  right  is  the  drawing  from  which  the  isometric  is  made ;  but 
the  necessary  dimensions  for  size  of  material,  distance  between  joists,  studding,  etc.,  have  been 
omitted  in  the  cut.  In  representing  incomplete  work,  as  in  this  plate,  it  is  necessary  that  the 
broken  lines  be  isometric,  as  otherwise  the  parts  will  appear  to  lie  in  different  planes.  Observe 
that  the  relation  between  the  different  surfaces  is  suggested  by  the  break  at  the  extreme  right 
of  the  isometric  drawing,  as  well  as  by  the  isometric  section. 

Fig.  12,  Plate  81,  illustrates  a  cube  drawn  by  a  system  which  is  a  modification  of  isometric 
projection.  The  effect  produced  is  more  nearly  that  of  a  perspective  drawing.  While  this  repre- 
sentation involves  a  little  more  labor  than  the  isometric  drawing,  it  avoids  much  of  the  distortion 
which  characterizes  the  latter.  The  system  is  particularly  well  adapted  to  illustrating  groups  of 
buildings  where  perspective  cannot  be  used  because  of  the  increased  difficulty  of  drawing,  and  the 
inability  to  measure  the  same.  It  is  also  suitable  for  Patent-Office  work. 

The  theory  upon  which  the  representation  is  made,  is  as  follows  :  A  cube  is  revolved  into  such 
a  position  that  two  of  the  axes,  or  edges,  as  C  G  and  C  B,  are  equally  inclined  to  the  plane  of 
projection,  while  the  third  axis,  C  D,  is  foreshortened  one-half  that  of  C  B  and  C  G.  The  angles 
which  the  edges  make  with  a  horizontal  line  are  7.2°,  41.4°,  and  90°.  This  involves  the  use  of 


OBLIQUE    PROJECTION.  97 

two  special  triangles,  one  of  which  must  have  a  right  angle  to  enable  the  vertical  lines  to  be  drawn. 
Two  scales  are  used:  a  full  scale  for  dimensions  parallel  to  C  B  and  C  G,  and  a  half  scale  for 
those  parallel  to  C  D.  As  in  isometric,  all  dimensions  must  be  made  parallel  to  one  of  the  three 
axes,  the  ellipse  of  the  front  face,  B  C  G  F,  may  be  described  by  obtaining  the  major  and  minor 
axes,  as  in  the  case  of  the  isometric  ellipse,  and  with  centres  on  these  axes,  describe  arcs 
tangent  to  the  edges.  The  axes  of  the  ellipse  on  face  D  0  G  H  do  not  coincide  with  the 
diagonals ;  but  a  very  close  approximation  to  them  may  be  -obtained  by  drawing  lines  K  L  and 
M  N,  respectively  perpendicular  and  parallel  to  C  B.  In  unimportant  work,  the  extremities  of  these 
axes  may  be  estimated  by  the  eye;  but  if  accuracy  is  required,  lines  K  L  and  M  N  must  be  drawn 
on  a  square  of  the  given  size,  and  their  intersection  with  the  circle  found,  as  was  done  with  the 
diagonals  in  isometric  projection.  Fig.  13  illustrates  a  chamfered  bolt-head  drawn  by  this  method, 
and  Fig.  14  represents  the  same  by  isometric  projection. 

OBLIQUE   PROJECTION. 

PLATE  31. 

In  this  system,  one  face  of  the  object  is  drawn  parallel  to  the  plane  of  projection,  and  all  edges 
perpendicular  to  this  face  are  drawn  as  oblique  lines.  Only  one  scale  is  commonly  used  for  the 
measurement  of  the  different  axes,  as  in  isometric  projection;  but  if  it  is  desired  to  foreshorten 
faces  perpendicular  to  the  front  face,  a  reduced  scale  may  be  employed.  Fig.  15  illustrates  a  cube 
of  equal  size  to  that  shown  in  Fig.  12,  drawn  in  oblique  projection.  The  axes  of  the  ellipse  on 
face  B  C  G  F  may  be  obtained  from  the  front  face  as  shown  by  the  dotted  lines.  The  curve  may 
be  drawn  by  circular  arcs  tangent  to  the  edges  of  the  parallelogram.  In  this  case  the  oblique 
lines  are  drawn  at  an  angle  of  30°;  but  45°,  or  any  other  angle,  might  equally  well  have  been  used. 

7 


98  OBLIQUE    PROJECTION. 

Fig.  16  illustrates  the  drawing  of  a  cabinet,  the  oblique  lines  being  drawn  at  an  angle  of  45°.  As 
the  top  projects  beyond  the  front  face,  it  lessens  the  apparent  height  of  that  face.  The  point  B 
on  the  upper  surface  of  the  top  of  the  case  is  directly  over  A,  the  upper  right-hand  corner  of  the 
front  face,  and  A  B  is  the  thickness  of  the  top.  B  D  and  C  D  indicates  the  amount  which  the 
top  overhangs  the  side  and  front,  respectively.  Oblique  projection  is  well  adapted  to  the  represen- 
tation of  this  class  of  work. 


€£Se 
OF  THE 
IVERS1TT 
i 


PLATE  1. 


PLATE  2 


PLATE  3. 


PLATE  1 


PLATE  4 


OF  THE 


PUVT5  5 


FIG.  37 


FIG.  36 


FIG.  39 


FIG.  40 


FIG..  41 


FIG.  42 


FIG.  43 


FIG.  44 


FIG,  45 


FIG.  46 1 


'- 


PLATE  4. 


PLATE  5. 


PLATE  6 


FiG.SO 


P FIG.  5  5 


D"   FIG.  57 


FIG.  58 


PLATE  7 


PLATE  8. 


PLATE  9. 


PLATE    9 


FIG.  71 


FIG.  72 


FiGV73 


>  74 


PLATE  10. 


PLATE  10 


ABCDEFGHIJKLMN 
OPQRSTUVWXY-Z& 


G  HIJKJL  M 

NOf>QrtSTUV\VXYZ 
abcdefg  hijklmnopq  rs  t 


123456739O?    1234567890^ 


PLATE  11 


PLATE  12. 


PLATE  12 


OBTAIN  SIDE 
VIEWOFPRISM 


PROS.  3 


OBTAI .  .  TOP  VIEW 
OFvJ-OGE. 


PROS 


PROS.  4 
DIAMETER  OF 
'CIRCUMSCRIBING 

CIRCLE   if*        . 


OBTAIN  FRONT  AND  SIDE  VIEWS 
OF  EQUILATERAL    TRIANGULAR 
PRISM    1\    HIGH  . 


OBTAIN  FRONT  AND  SIDE 
OF  EQlll LATERAL  TRIANGULAR 
PYR/^VIID     2~HIGH. 


OBTAIN  TOP  VIEW  OF 

RECTANGULAR   PYRAMID. 


OBTAIN  FRONT  AMP  SIDE  VIEW 
OF  PENTAGONAL  PRISM. 
2JJH1GH 


PHOB.  7 


Pnoe.  8 

TOP  AND  SIDE  VIEWS 
OF  HEXAGONAL  PYRAMID. 


OBTAIN  FRONT  AND  SIDE 


PROS.   9 


OBTAIN  TOP  VIEW. 


PROS.  10 

COPY  TOP  VIEW  FROM  PRECED- 
ING PROBLEM  BUT  WITH  THE 
LONG  EDGES  AT  AN  ANGLE  OF 
30°WITH  THE  FRONT  AXIS. 
OBTAIN  FRONT  AND  SIDE  VIEWS. 


FlG.1 


PROB.    13 
OBTAIN  TOP  AND  SIDE  VIEW. 

FlG.2 


PROB.  1 1 

OBTAIN  TOP  VIEW  WITHOUT 
THE  USE  OF  AXES 


PROB.  12 

COPY  TOP  VIEW  FROM  PRECEDING 
PROBLEM  BUT  WITH  LONG  EDGES 
MAKING  AN  ANGLE  OF  3O°WITH 
THE  FRONT  AXIS. 
USE  AXES  BUT  OMIT  PROJEC- 
TION LINES. 
OBTAIN  FRONT  AND  SIDE  VIEWS 


OBTAIN  TOP  AND  FRONT  VIEWS  • 


FlG.3 


FIG.  4 
OBTAIN  FRONT  ^ND  SIDE  VIEWS 


PLATE  13. 


PLATE  13 


FIG.  4          A >JT B  -FIG.  5 


PLATE  14. 


PLATE  14 


PROB.  14 

PROB.  15 

OBTAIN 

TOP  VIEW. 

U  ,6  j 

L»  i    o 

~|~>j<^: 

REVOLVE  THE  OBJECT 

REVOLVE  THE  OBJECT 

*  —  '5  X 

K  i  y 

REVOLVE  THE  OBJECT 
AS  SHOWN  IN  FIG.1 

!           L 

AS  SHOWN  INFIG.1 
30*  ABOUT  SIDE  AXIS 

AS  SHOWN  IN  FIG.2 
25*  ABOUT  VERTICAL 

T 

Dfc. 

i 

30°  ABOUT  SIDE  AXIS  • 

A       A  ^ 

TO  LEFT  . 

AXIS. 

I     — 

/  \    j\^ 

FlG.1 

FIG.  2 

k    ib    3            FiG.1 

.     FIG.  2 

FIG.  3 

REVOLVE  THE  OBJECT 

REVOLVE  THE  OBJECT 

REVOLVE  THE  OBJECT 

REVOLVE  THE  OBJECT 

HEVOLVE  THE  OBJECT 

AS  SHOWN  INFIG.1 

A8  SHOWN  INFIG.1 

AS  SHOWN  IN  FIG.1 

AS  SHOWN  IN  FIG.2 

AS  SHOWN  INFIG.S 

30°  ABOUT  FRONT  AXIS. 

60*  ABOUT  VERTICAL  AXIS. 

20°  ABOUT  FRONT  AXIS 

15°  ABOUT  FRONT  AXIS 

3S°ABOUT  VERTICAL 

FORWARD  . 

FORWARD  . 

AXIS. 

FiG.3 

FlG.4 

FIG.  4 

FlG.5 

FIG.  6 

7 

PROB.  16 

PROB.  17 

i    B>- 

T^ 

«    / 

x^S^ 

REVOLVE  THE  SURFACE 

REVOLVE  THE  SURFACE 

i    / 

/ 

I            \  \w 

AS  SHOWN  INFIG.1 

AS  SHOWN  IN  FIG.2 

Jf    i    '     1           *          \,                    REVOLVE  CONE  3O  "ABOUT 
E?~|~               ^F^/^                 VERTICAL  AXIS. 

f\lL 

3O°  ABOUT  SIDE  AXIS 
TO  LEFT  . 

25°  ABOUT  VERTICAL 
AXIS. 

\j 

//                                   OBTAIN  3  VIEWS. 
_x/^ 

AT*- 

/           \1>    /      ^r. 

K-ti—  >l            FIG.  1 

FIG.  2 

FIG.  3 

tj-r" 

^/f* 
},*/       / 

REVOLVE  THE  SURFACE 

REVOLVE  THE  SURFACE 

REVOLVE  THESURFACE 

40& 

AS  SHOWN  INFIG.1 

AS  SHOWN  IN  FIG.  2 

AS  SHOWN  INFIG.S 

20"  ABOUT  FRONT  AXIS 

15°  ABOUT  FRONT  AXIS. 

35°  ABOUT  VERTICAL 

y\ 

f\        / 
J     \     / 

FORWARD. 

AXIS. 

«^—  3^ 

rv  I 

FIG.  4 

FIG.  5 

FlG.6 

PLATE  15. 


PLATE  15 


FIG.  1 


FIG.  5 


\ 


\ 


FIG.  6    \  / 

\L' 


FIG.  8 


PLATE  16. 


PLATE  1  6 


PLATE  17. 


PLATE  17 


4?      .    PROB.  1 
\ 

^  TOP!  VIEW  OF  \. 

—* r-  -       -»-  RCVOLVED 

HEXAGONAL  PRISM.        V 

SECTION 


DEVELOPMENT 


PROB.  2O 


REVOLVED 
SECTION 


DEVELOPMENT 


TOP  view  OF 

SQUARE  PYRAMID. 


21 


REVOLVED 
SECTION 


DEVELOPMENT 


*       PRoa  22 


1  SIDE   VIEW 


f    PROB.23 


SIDE    VIEW 


PROB.  24 

iON 

INSCRIBED 
IN    2£*CIRCLE. 
REVOLVED 
SECTION 


PROB.  25 


DEVELOPMENT 


DEVELOPMENT 


DEVELOPMENT 


PLATE  18. 


PLATE  1 8 


FIG.  4 


PLATE 


PLATE   19 


;  20. 


DEVELOPMENT  OF  CYLINDER   A 


DEVELOr  ,*ENT  OF  CYLINDER   B 


PLATE  20. 


PLATE  2O 


PLATE  21 


PLATE  21 


PROS.  31 


DEVELOPMENT  OF 
CYLINDER  A 


PROB.  3J 


DEVELOPMENT  OF 
CYLINDER    A 


PROS.  33 


f  PROB.  35 


DEVELOPMENT    OF  EQUILATERAL 
TRIANGULAR   PRISM 


DEVELOPMENT  OF  CYLINDER 


PROB.  34 


DEVELOPMENT  OF  EQUILATERAL 
TRIANGULAR   PRISM 


DEVELOPMENT  OF    HEXAGONAL   PRISM 


DEVELOPMENT  OF  HEXAGONAL 
PRISM 


PROB.  36 


TOP    VIEW 


DEVELOPMENT  OF  HEXAGONAL 
PRISM  A 


PLATE  22 


FIG.  2 


FIG.  3 


\ 

T 

t 

i 


OF  THE 

UNIVERSITY 


PLATE  22. 


PLATE  23. 


PLATE  24. 


PLATE  24 


EQUABLE    |SPI RJAL. 


"  PITCH 


INVOLUTE  OF 

-1-f  SQUARE 


[PROB.  37 
INVOLUTE   OF 


EQUILATERAL    TRIANGLE. 
SIDES   ^ 


PROB.  38 


SQUARE  THREAD    1  £ P 


•s-f* 


THREAD 
t|  PfTCt 


J&BQB.  39 


SECTION  OF 

NUT  FOR  I 
SQUARE  | 
"  "JTHREJAD" " 
1 1^' PITCH 

«          J. 
•  —        —  1b 


V    THREAD    [IJ  PITCH 
ANGLE   OFiV    45* 


RIGHT 
HAND  THREAD 


R.  H.  DOUBLE 


PROB.  4O 


V  THREAD  t;"p  L.H.  V  TH.-f"p. 


UH.SQ.       |'P      "«  R.H.  SQ.TKj"p- 


PROB.  41 


.1-!. SINGLE  RH  DOUBLE  L.H. SINGLE 


PLATE  25. 


PLATE  25 


PLATE  26. 


PLATE  26 


^         PITCH 


PLATE  27. 


PLATE  27 


UPITCHJ 


/ 


FIG.  5 


FIG.  9 


PLATE  28. 


PLATE  28 


PLATE  29. 


PLATE  29 


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5 

PLATE  30. 


PLATE  30 


PLATE  32 


PLATE  32. 


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